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hello and welcome to college physics 1
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lecture 14 newton's laws and free body
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diagrams
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in our last lecture we introduced force
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and different types of forces
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in this lecture our goal is to go
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through newton's three laws of motion
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and then introduce how we represent each
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of these forces we've discussed
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visually with free body diagrams
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to begin let's recap newton's first law
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because we did introduce this in our
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past lecture
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in general uh when we think of newton's
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first law we're considering any object
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that has no forces acting on it none
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whatsoever
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so in theory this is fairly
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unrealistic on earth because almost
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everything has a force acting on it
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especially because of weight
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but that said if we consider an object
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that has no forces acting on it
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whatsoever
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then we can talk about newton's first
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law
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if the object is at rest
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with no forces acting on it
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it will remain at rest
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i mean this makes sense saying you leave
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a pencil sitting on your tabletop it's
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not going to move unless something acts
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to move it
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and then if the object is moving with no
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forces acting on it it will continue to
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move in that same direction in other
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words a straight line and without
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changing its speed it'll stay the same
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constant speed until
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something acts on it
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now again this is somewhat hard to
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visualize on earth because it's not
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realistic
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so that takes us to space
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in many cases the best example of this
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is an object moving through
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space
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so on the top i just put in an image
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from south park of kenny uh floating
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through space but notice he's moving in
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a straight line and perhaps somewhat
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hard to tell but at a constant speed
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and then another great example is this
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dog
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that was in orbit or on the earth so
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effectively experiencing zero g's
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and as a result in each of the frames of
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this animation you can see the dog
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moving in a straight line and at a
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constant speed even when he's paddling
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in one of the frames it's not changing
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his motion at all
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so
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this is newton's first law again this is
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just a recap
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so this brings us then to something new
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here we're going to consider an
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experiment that we can't really do well
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or easily on earth
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in my class in person
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we can use an air track that is a
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surface like an air hockey table where
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there's holes in it and air blown into
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the surface
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so that's anything laying on top of the
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surface can
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experience almost no friction but even
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still it's not a perfect experiment
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so just consider a block
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on a frictionless surface
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and then imagine you take a rubber band
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put it around the block
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and then you pull it
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specifically you pull it to maintain a
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constant force in other words you aren't
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letting the rubber band compress or
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stretch at all so it's staying at a
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constant force or a constant
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stretchiness
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if we do this experiment successfully we
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would find a number of different things
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first of all and this is the hardest one
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to visualize
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the block pulled with a constant force
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in other words not letting your rubber
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band stretch or compress
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would have to move with a constant
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acceleration
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meaning
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you would have to constantly pick up
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speed
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and again even if we had the air track
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and i showed you it it's hard to
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visualize it because you aren't the one
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actually moving the block you'd have to
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feel how stretchy the rubber band is but
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just try to visualize that
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so if you want to move something with a
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constant force you would have to
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constantly accelerate it
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the other two results of this experiment
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are fairly straightforward
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the acceleration of the object would
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increase if you increase the force
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in other words if you pull harder the
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block would accelerate faster
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and
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the acceleration would decrease
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if the mass of the block increased
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so just imagine you try to apply the
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same force to an object that's 10 times
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heavier
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well it's going to be harder to
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accelerate that object
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so what we can do with this information
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is sort of combine it all into one big
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piece of information
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the first statement statement we made
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was that with a constant force you have
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a constant acceleration
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and the acceleration will increase if
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the force does
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so that's telling us that there is a
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direct relationship between force and
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acceleration
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if force increases so does acceleration
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and then we say acceleration decreases
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if the mass increases in other words
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acceleration and mass are inversely
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proportional to one another
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putting all of that together we arrive
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at newton's second law
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newton's second law is a mathematical
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equation
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it states an object of mass m
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subjected to forces be it 1 2 3 and so
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on
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will undergo an acceleration given by
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f net
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over m
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where f net is the net or total force
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and m is the mass
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so recognize on this left-hand equation
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that
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we see a direct relationship between a
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and f we said that if you increase the
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net force you increase the acceleration
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but if you increase the mass in other
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words you increase m underneath this
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fraction
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well then you're dividing by more which
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means your acceleration would decrease
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so this equation
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is what we just stated in our experiment
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it's just a formulation to show it
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now this rearranges of course to one of
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the most famous equations in all of
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physics
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f equals m a
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i'd argue that this and probably e
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equals m c squared are like the two most
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well-known equations in physics
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um and for good reason this is going to
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be an equation that we use throughout
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the next
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i don't know
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seven lectures probably um it's the
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premise of all the problems we're going
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to work out uh and we'll actually see
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that an equation this small and innocent
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looking is not so much that case as we
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get into this further
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now the only thing left to discuss on
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this
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equation is the units
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up until this point we've introduced
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what forces but we never said what the
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units of force are
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this equation allows us to visualize
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that
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we see f is equal to m a
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m or mass is measured in physics
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in kilograms
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a or acceleration is measured in meters
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per second squared
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so combine the two you have a kilogram
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times a meter per second squared
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that is the unit of force
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a kilogram meter per second squared
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but that's kind of ugly and we don't
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like to work with that most of the time
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so we just give it a new name
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we call a kilogram meter per second
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squared a newton or 1n
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this is what we use as the unit of force
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the newton
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all right this then brings us to
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newton's third law of motion out of
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three newton's third law is an action
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reaction discussion
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so consider for example you are
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hammering a nail into a wall
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obviously you with the hammer are
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applying a force to the nail so that it
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is put punched into the wall
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but the nail also pushes back at the
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hammer in fact you experience that as a
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recoil
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right if you hit something really hard
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with a hammer you can feel that recoil
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in the handle
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an even better example perhaps is if you
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fire a rifle or a shotgun with the stock
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of that weapon up against your shoulder
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when you fire the bullet or the shell
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out the front the weapon kicks back into
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your shoulder
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you can experience that reaction
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so in general we don't have to be
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specific and say a hammer or a gun we
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can just say if object a exerts a force
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on some object b then object b will
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exert a force back on a
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this is a pair of forces that we call an
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action reaction pair
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now i will note i don't necessarily at a
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personal level like that term action
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reaction
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mostly just because by definition
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reaction means
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you see or experience something and then
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respond
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these forces occur simultaneously so
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it's not really that one happens and
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then the other one decides hey i need to
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apply a force too they happen at the
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exact same time
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but regardless well i mean thinking of
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it as action reaction is just a really
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easy way to remember what's happening so
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it is useful
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so newton's third law says in general
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that every single force
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occurs as a member of this
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action-reaction pair of forces
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and they have equal magnitude so it's an
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equal force between the two
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but they are pointed in opposite
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directions
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so you commonly hear this is for every
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action there's an equal and opposite
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reaction that's a common statement
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so with this we now have a basic
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understanding of what forces are and the
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three laws of motion that govern them
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these laws apply to any object anywhere
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anywhere in the universe
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and what's kind of crazy to think about
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is that this newton's third law this
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action reaction it's true for every
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single force
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it's even true for the gravity of earth
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holding the moon in orbit the moon is
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also tugging on the earth with an equal
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force the difference is the moon is so
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much smaller that it orbits around the
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earth instead of vice versa
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but i digress
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to conclude the notes part of this
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lecture we have to introduce free body
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diagrams
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now if you have my actual course you
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will hear me say this a lot but these
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are severely important you really want
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to understand these
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so
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in the next slide here i'm just going to
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step you through what a free body
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diagram is and how to draw them
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and then for the remainder of this
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lecture we're just going to work out
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problems like questions together
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so you can get some practice working
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with these free body diagrams before we
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try drawing them ourselves
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so
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in general a free body diagram or as i
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abbreviate them fbd
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these are diagrams that represent
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objects as a particle or dot
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and then show all of the forces acting
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on the object as vectors or arrows
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pointing away from the dot
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so you'll see a lot more about what i
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mean here in just a few minutes but
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let's run through the steps the single
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most important step to all of this
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is identifying the forces that are
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acting on your object
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if you don't know what forces are acting
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on your object you can't draw one of
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these diagrams properly and then you
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certainly couldn't work out the problem
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correctly
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so really everything relies on you
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understanding the problem you have
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and then the forces involved within it
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so identify your forces
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steps two and three kind of go together
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and they're hardly steps at all
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for every free body diagram you draw a
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coordinate system in other words an x
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and a y axis and if you're along a ramp
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like we discussed in the past you tilt
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your axes along the incline so that the
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x-axis is along the slope
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so i mean really these all start by you
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just drawing basically a large plus sign
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on your paper your coordinate axes
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don't forget to label them and then step
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three again barely even a step just draw
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a big dot at the center
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so at this point all you've done is
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drawn a big plus sign and a little dot
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in the middle
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now once you've identified your forces
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you'll be able to draw them onto your
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free body diagram
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so draw vectors in other words arrows
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representing each of the forces that
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you've identified
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last but not least and
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i tend to forget this one myself or just
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not do it i'd say that it's optional
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i would still like you to do this but
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you want to draw and label the net force
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vector afterward
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you typically don't draw this directly
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onto the axes like you do with the other
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forces though
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a lot of times you'll just kind of draw
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this off to the side
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just so that you
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for your own sake
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can remember what direction the net
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force will be pointing if there is one
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and i suppose i should add into step
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number four and it's very important that
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you label each of your vectors
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otherwise it's just a bunch of arrows on
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a page
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so labeling them appropriately is very
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important as well especially when it
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comes to me grading your diagrams
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because
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like i mentioned if there's no labels i
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don't know what force you're thinking of
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and it's just an arrow on the page
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so don't forget to label your vectors
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okay so this isn't entirely helpful
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since you haven't even seen one of these
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drawn yet so let's step through a few
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questions and after several of them we
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will get to free body diagrams
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so
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we're going to go through a lot of
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questions here i believe there's
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actually 12 of them
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so
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i encourage you to pay attention to each
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one as they are very important it'll get
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you some practice with these diagrams
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toward the end as well
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so question one
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a cart is pulled to the right as shown
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below at a uh with a constant force
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it asks how will its acceleration graph
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look
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so this is a throwback to graphs take a
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moment to think about your answer and
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then come back when you think you have
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it
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okay
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well in this case the key here is that
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it's being pulled with a constant force
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generally what you want to do is think
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of newton's second law here which says f
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net
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is equal to m a
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so with this in mind
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it says that there's a constant force
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being applied so your force f is not
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changing
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well we know mass isn't changing
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so if neither f or m are changing well
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then how can a change
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right
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if a net force is constant the
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acceleration also has to be constant
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so on an acceleration graph
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that would show up as a straight
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horizontal line like we see here
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in c
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a constant force means you're you'll be
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applying a constant acceleration to the
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object
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all right
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question two
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a constant force causes an object to
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accelerate at four meters per second
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squared
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what is the acceleration of an object
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with twice the mass but the same force
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i'll give you a hint it has to do with f
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equals ma
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okay
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the problem here is asking about the
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acceleration so let me rearrange our
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equation
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a
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equal f over m
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equal to f over m
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well what we've done is in this problem
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it says that we are going to
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double the uh mass right
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so it says it has twice the mass so what
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we're basically doing is
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doubling the mass so multiply by two
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down here but it has the same force so
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one
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so we're keeping the same force we're
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doubling the mass
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so our acceleration is going to change
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by a factor of one half
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in other words if we started out with
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four meters per second squared we would
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end up with an answer
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of two meters per second squared
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and again this holds true to ride our
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idea that if you increase the mass
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you're going to decrease
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the acceleration
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all right let's do one more like the
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last
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an object when pushed with a net force f
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has an acceleration of 2 meters per
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second squared
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now
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you double the force and you quadruple
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the mass
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its acceleration will be what
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all right
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well again this problem is asking about
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acceleration so let me rearrange the
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equation again
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for f over m
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similar to what we did last time let's
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think about by what factor we are
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changing things we are saying that the
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force is twice as large
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so
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you're changing the force by a factor of
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two
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well we're also quadrupling the mass or
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multiplying it by four times
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so
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we've changed our acceleration by a
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factor of two over four which is one
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half
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so we
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have our
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acceleration once again
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so the answer
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half of two meters per second squared
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is one meter per second squared
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all right well let's step away from the
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math for the rest of our questions let's
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think conceptually and then visually as
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well
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the next two questions are very good i
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like these ones a lot
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this question says a 40 car train
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travels along a straight track at 40
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miles per hour
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separately
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a skier speeds up as she skis downhill
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based on these two different situations
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on which could we say that the net force
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is greater if either
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at all
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a lot of times with this question my
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students will just have a big question
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mark above their head
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uh it's weird to think about i mean a
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train and a skier and how can you
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compare their forces
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well i mean you can you actually can
00:19:09
tell there's a key here
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the skier is speeding up
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the train is moving at 40 miles per hour
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the train's moving at a constant speed
00:19:21
the skier is accelerating
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again think of f equals ma
00:19:28
to have a force a net force right
00:19:31
we need to be
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accelerating
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if you don't have an acceleration you do
00:19:38
not have a net force
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so the train even though it's this giant
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massive object has no net force acting
00:19:45
on it
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but the skier even though we don't have
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a number
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does have an acceleration we know
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there's an acceleration because she is
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speeding up
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so we don't know how big the net force
00:19:58
is but we know it's more than zero
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so the skier
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has the greater net force
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okay so again the train does not have a
00:20:08
net force because it is not accelerating
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the skier is accelerating so they do
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have a net force
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so the answer is b
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all right let's move on to question five
00:20:25
an object on a rope is lowered at a
00:20:28
constant speed
00:20:29
which of the following is true
00:20:36
okay
00:20:37
well here we should be able to recognize
00:20:39
what forces are acting on our object
00:20:42
we should note that there is a weight
00:20:45
pulling downward on the object
00:20:47
right we have a weight pulling down
00:20:50
and there is a tension holding this
00:20:52
object upward
00:20:54
so we know that there's a weight and a
00:20:56
tension
00:20:58
now
00:20:58
the key here is it says it's at a
00:21:00
constant speed as we just stated on our
00:21:03
last question by f equals m a
00:21:07
we should recognize that
00:21:09
in order to have a net force you would
00:21:11
have to have an acceleration
00:21:14
we don't accelerate in this problem so
00:21:16
we do not have a net force
00:21:19
right no acceleration
00:21:21
no net force
00:21:23
which means
00:21:25
even though there are forces there can't
00:21:27
be a net force which must mean
00:21:31
the tension pointing up has to equal the
00:21:33
weight pointing down so that there is no
00:21:36
net force
00:21:38
in other words one of them isn't pulling
00:21:40
harder
00:21:41
in the other direction
00:21:43
so the answer is that they are equal
00:21:46
our answer is b
00:21:51
okay
00:21:54
question six
00:21:55
a very similar one the only difference
00:21:58
is
00:21:58
now instead of moving at a constant
00:22:00
speed it says that it is lowered at a
00:22:03
decreasing speed
00:22:06
which is true
00:22:15
okay
00:22:16
well the situation is the same we have a
00:22:18
weight pointing downward so i'll draw
00:22:20
that
00:22:22
here's our weight
00:22:25
upward is our tension
00:22:26
but here is the key and this is
00:22:28
important be careful with how i or pay
00:22:30
attention to how i say this
00:22:33
it is lowered
00:22:35
at a decreasing speed
00:22:38
from past lectures we learned how to
00:22:40
deal with the direction of acceleration
00:22:44
recall that if an object is slowing down
00:22:47
like the object here
00:22:49
acceleration points in the opposite
00:22:51
direction of the object's motion
00:22:55
so if our object is pointing downward
00:22:58
or moving downward which it is
00:23:00
then our acceleration because it's
00:23:02
slowing down must point in the opposite
00:23:03
direction or upward
00:23:07
so if we have
00:23:10
an upward pointing acceleration
00:23:14
then we must also have an upward
00:23:16
pointing net force
00:23:18
if a points up
00:23:20
net force points up
00:23:22
which means we have to have a larger
00:23:24
force that points upward
00:23:27
so i'll draw that as a longer arrow for
00:23:30
tension
00:23:32
so tension in this case is greater than
00:23:35
the weight of the object
00:23:36
so the answer
00:23:38
is a
00:23:40
so again be careful this is not very
00:23:43
easy for a lot of students
00:23:45
we are slowing down while moving
00:23:47
downward which means because we're
00:23:49
slowing down acceleration points in the
00:23:50
opposite direction or upward
00:23:53
and based on our concept of the
00:23:55
relationship between f and a
00:23:58
if acceleration is pointing upward so
00:24:00
does the net force
00:24:02
so whichever force is pointing upward
00:24:04
also
00:24:05
or uh has to be the larger force
00:24:08
that's the tension force so the answer
00:24:11
is a
00:24:15
all right
00:24:16
on to my favorite question out of the
00:24:18
bunch
00:24:19
a mosquito runs head-on into a truck
00:24:22
splat
00:24:24
which of the following is true
00:24:25
during the collision
00:24:30
in this case we have to think about
00:24:33
newton's
00:24:34
third law
00:24:36
newton's third law states that every
00:24:38
force occurs as an action reaction pair
00:24:41
where the forces are in opposite
00:24:43
directions
00:24:44
but are equal in magnitude
00:24:49
equal in magnitude
00:24:51
which means
00:24:52
perhaps surprisingly to some people
00:24:55
the answer is c
00:24:56
they will exert the same force on each
00:24:58
other the mosquito to the truck and the
00:25:00
truck to the mosquito
00:25:02
now this is very odd to some people so
00:25:05
let me just
00:25:06
show efficient again
00:25:12
in the case of let's say the truck
00:25:13
hitting the mosquito
00:25:15
well think about what's going on here
00:25:17
the truck has a really large mass
00:25:22
but
00:25:23
because the truck hits the mosquito is
00:25:25
it going to change its speed very much
00:25:27
i mean no the mosquito's not going to
00:25:28
slow the truck down at all so it has a
00:25:31
very
00:25:32
low
00:25:33
acceleration
00:25:36
on the other hand the mosquito has a
00:25:38
really low mass
00:25:41
i mean tiny mess
00:25:43
but it's just floating around around
00:25:45
having its you know day trying to ruin
00:25:47
people's day and then it's suddenly hit
00:25:49
by a truck moving i don't know 70 miles
00:25:51
per hour so it's going to be suddenly
00:25:53
accelerated
00:25:54
really quickly so it's got a low mass
00:25:57
and a high acceleration but the truck
00:25:59
has a high mass and a low acceleration
00:26:02
they are equal
00:26:06
all right
00:26:07
let us now practice free body diagrams
00:26:10
now that we have this background
00:26:12
let's try drawing them
00:26:15
the first of five of these diagram
00:26:18
questions
00:26:19
says an elevator which is lifted by a
00:26:21
cable is moving upward and slowing down
00:26:25
which is the correct diagram
00:26:32
all right well it's moving up and
00:26:34
slowing down we know there are two
00:26:36
forces in this problem weight pointing
00:26:38
downward
00:26:40
tension in the cable pointing upward
00:26:42
like d shows there is not a third force
00:26:45
it says f elevator
00:26:47
the force pointing upward is caused by
00:26:50
the tension so there isn't an extra
00:26:51
force here
00:26:53
so that's irrelevant
00:26:55
it is moving upward but motion is not a
00:26:58
force
00:27:01
and because it is slowing down that
00:27:03
means there is an acceleration which
00:27:05
means there is a net force
00:27:08
we can rule out option d and e as a
00:27:10
result now
00:27:16
so
00:27:17
keeping in mind what we said in our last
00:27:19
question
00:27:21
we have a situation where we're moving
00:27:24
up but slowing down
00:27:26
slowing down means acceleration points
00:27:28
in the opposite direction of motion
00:27:30
which would be downward
00:27:33
so if we have an acceleration that's
00:27:35
pointing downward
00:27:36
we have to have a net force that is
00:27:38
pointing downward as well
00:27:41
in other words whichever force is
00:27:42
pointing downward has to be the larger
00:27:44
force
00:27:45
leaving us with the answer of c
00:27:49
weight down tension up but because
00:27:51
acceleration points downward the force
00:27:54
that's pointing downward has to be
00:27:55
larger to give us the larger net force
00:28:01
all right so that's a chain
00:28:03
of thought processes you have to go
00:28:04
through to get there so be careful with
00:28:07
these
00:28:09
question nine
00:28:11
a ball has been tossed straight up
00:28:14
which of the following is the correct
00:28:15
free body diagram just after the ball
00:28:18
has left the hand
00:28:19
as always
00:28:20
ignore air resistance
00:28:26
in this case
00:28:28
the ball has left the hand
00:28:30
meaning there are no contact forces on
00:28:32
the ball nothing is physically touching
00:28:35
it
00:28:36
what that means is there's only
00:28:38
one force acting on it
00:28:41
it's weight the weight due to gravity as
00:28:44
it moves through the air
00:28:46
the answer is d
00:28:47
any object moving through the air
00:28:50
will only ever have one force acting on
00:28:53
it the weight
00:28:55
assuming we're ignoring air resistance
00:28:56
of course
00:28:59
so the answer is d only the weight is
00:29:01
acting on the ball
00:29:03
if the ball was still in the person's
00:29:05
hand as it was being pushed upward
00:29:08
then you'd certainly have an upward
00:29:09
force but we don't in this case
00:29:16
next question
00:29:18
a ball is hanging from the ceiling by a
00:29:20
string
00:29:21
it is pulled back and released
00:29:23
which is the correct free or which
00:29:25
diagram below is the correct one just
00:29:27
after the ball has been released like
00:29:29
you can see on the image on the top
00:29:30
right
00:29:35
all right
00:29:37
in this case
00:29:38
the first thing i'll mention is that
00:29:41
the rope is at an angle
00:29:43
but it's not an object moving on a ramp
00:29:46
or an incline so we do not tilt our axes
00:29:49
in this case like shown in c or excuse
00:29:52
me like shown in e
00:29:54
we don't tilt our axes that is only for
00:29:56
objects that are moving along an
00:29:58
inclined surface this is just a ball
00:30:00
swinging back and forth
00:30:04
let's think about one of the forces
00:30:06
acting on this ball one of the forces is
00:30:09
weight
00:30:10
by definition weight always points
00:30:13
vertically downward
00:30:15
which means we can rule out
00:30:17
d
00:30:18
d does not show the weight force
00:30:20
pointing downward so that is incorrect
00:30:24
a b and c all show that
00:30:27
so a shows that there's only the weight
00:30:30
but
00:30:31
there's more than that
00:30:34
the object is being held up by a rope
00:30:37
so there should be a tension force
00:30:40
that tension force by definition always
00:30:42
points in the direction of your rope
00:30:43
which is up and to the right
00:30:45
both b and c show that
00:30:49
at this point we just have to ask
00:30:50
ourselves is there a third force
00:30:54
c shows the third force pointing down
00:30:56
into the right which is
00:30:58
the direction of the well not quite it
00:31:01
is the direction of the object's motion
00:31:04
but
00:31:05
motion is not a force
00:31:08
nothing is physically pushing the ball
00:31:11
down into the right
00:31:12
there is no third force
00:31:16
just to give you an idea on why it does
00:31:18
move that way
00:31:20
think of vector addition
00:31:22
right if you can just visualize this say
00:31:24
someone's pulling down on the ball here
00:31:26
someone's pulling up on the right to the
00:31:27
ball there that means the ball is going
00:31:29
to move with a net force that points in
00:31:32
that direction
00:31:34
right those two forces combined are
00:31:36
going to pull the object down into the
00:31:37
right which is why we see the ball doing
00:31:39
that
00:31:41
anyways this means the answer is
00:31:44
b you have the weight of the ball acting
00:31:46
downward and the tension in the row
00:31:48
pointing up and to the right
00:31:53
question 11
00:31:55
a car is parked on a hill
00:31:58
which is the correct free body diagram
00:32:04
well in this case we are on an incline
00:32:06
so we do tilt our axes which all of
00:32:08
these show
00:32:10
what we need to do though is start
00:32:12
listing our forces first of all like any
00:32:15
problem on earth we have a weight
00:32:18
we know that weight
00:32:20
always points vertically downward
00:32:23
well
00:32:24
d does not show a force pointing
00:32:27
straight down
00:32:29
d is incorrect because of that
00:32:31
a b c and e all show a weight force
00:32:35
pointing straight down
00:32:38
let's think of another force
00:32:41
car is parked on a hill
00:32:44
it's parked meaning it's not moving
00:32:47
so something has to be resisting its
00:32:49
motion and that's friction
00:32:52
friction is acting to resist the motion
00:32:54
of this car the car wants to move down
00:32:57
the hill
00:32:58
so there must be a friction force
00:33:00
pointing
00:33:01
up the hill
00:33:03
a
00:33:04
does not show a force pointing up the
00:33:06
hill in the x direction
00:33:08
b
00:33:09
does not show a force pointing up the
00:33:11
hill it shows a force pointing down the
00:33:13
hill which is not correct
00:33:15
both c and e show the weight pointing
00:33:18
downward and a friction force pointing
00:33:20
up the hill
00:33:22
so we now at this point ask ourselves is
00:33:24
there a third force
00:33:27
in this case
00:33:29
there is
00:33:31
there is a third force the answer is c
00:33:34
here
00:33:35
that third force is because the car is
00:33:37
on a surface
00:33:39
a surface always pushes back
00:33:41
perpendicularly by the normal force
00:33:45
so that's what we're seeing here we're
00:33:46
seeing a normal force pointing off the
00:33:48
surface of the ramp
00:33:49
a static friction force holding the car
00:33:52
in place
00:33:54
and the weight of the car pulling it
00:33:56
downward
00:33:57
c is the correct
00:33:59
free body diagram
00:34:03
this brings us to our very last question
00:34:06
there's been a lot of them
00:34:08
question 12.
00:34:09
a car is towed by a rope to the right at
00:34:12
a constant speed
00:34:13
which diagram is correct
00:34:19
okay i'm going to go through this one a
00:34:21
little quicker since we should have some
00:34:22
practice
00:34:23
we know there's a weight pulling the car
00:34:25
downward all of them show that
00:34:28
we know there is a normal force pulling
00:34:30
or pushing upward
00:34:32
all of them show that
00:34:34
there should be
00:34:35
a tension force in the cable pulling it
00:34:38
forward to the right
00:34:41
a does not show a force to the right
00:34:44
but there should also be a force
00:34:46
resisting the car's motion
00:34:48
it doesn't say that it's frictionless so
00:34:50
there should be a friction force
00:34:52
resisting the car's motion as well
00:34:56
against the up against the direction of
00:34:57
the car so to the left
00:35:00
so there isn't a force to the left in b
00:35:02
we can rule that one out as well
00:35:05
this brings us to options c d and e
00:35:08
all of these show the correct four
00:35:10
forces
00:35:11
but there's a difference
00:35:13
c shows that the tension is greater than
00:35:16
the friction
00:35:18
uh oops that's normal force this is
00:35:21
weight
00:35:22
um imagine using the same letters over
00:35:24
here
00:35:25
in d the two forces friction and tension
00:35:27
are the same
00:35:29
and in e the friction is the larger of
00:35:31
the two
00:35:33
well keep in mind it says that this is
00:35:35
moving at a constant speed
00:35:38
that means there is no net force
00:35:42
so all of the forces should be equal and
00:35:44
opposite
00:35:45
and that's what d shows
00:35:48
d shows our equal and opposite forces
00:35:51
so that is our answer
00:35:56
all right so uh at this point we've at
00:35:58
least introduced what a free body
00:36:00
diagram is now and how to draw them
00:36:03
so in the future we will actually be
00:36:05
drawing them ourselves from scratch not
00:36:06
just picking them from a multiple choice
00:36:08
question
00:36:09
so hopefully at least for now this has
00:36:11
given you at least
00:36:12
a basic understanding of what they look
00:36:14
like and how they will be drawn
00:36:16
in our next lecture we will start to add
00:36:18
some math into all of this and start
00:36:20
drawing these ourselves but until then
00:36:23
thanks for watching and have a great day
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