WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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We are going to calculate the force on
each of the feet of this horse
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and it's going to be a force that is
somewhat upwards
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to balance the weight of
the horse and rider
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and it's also going to be
somewhat to the left
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to balance the force exerted by
the wall to the right
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and so it's gonna have some angle *Θ* here
above horizontal towards the wall.
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So the first condition of equilibrium
is to say that the total
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forces in the x-direction and in
the y-direction equal zero
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or you could say that the forces to the left
equal the forces to the right.
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So this is 2 times because
there are two feet
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2 times the force on a single foot
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multiplied by *cosine Θ* to give us this
component which is horizontal here
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that equals the force exerted
by the wall to the right.
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And then we can talk about the y-direction
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and say that the vertical component of
the force exerted by a foot
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multiplied by 2 equals the total
weight downwards, *mg*.
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And there are three unknowns here:
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we don't know the force exerted by the wall
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nor *Θ* nor do we know the force
on a single foot
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and so we need to have three equations to
solve three unknowns
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and so far we have two equations
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and then we'll use the second
condition of equilibrium
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which says that the total clockwise torque
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has to balance the total counter clockwise
torque to find equation (3).
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So we have the force exerted by the wall
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multiplied by its perpendicular distance to
the pivot at the foot—1.20 meters—
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equals the counter-clockwise torque
due to the weight
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which is 0.35 meters perpendicular distance
to the pivot.
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So we divide both sides by 1.20 meters here
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and this gives us the force
exerted by the wall
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*mg* times 0.35 divided by 1.20
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and then we can do some substituting
and rearranging
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to eventually solve for some of
the unknowns.
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So returning back to equation (1),
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we can solve it for the force
exerted on a foot
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and it's going to be the force exerted by
the wall divided by 2*cos Θ*
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when you divide both sides by
2*cos Θ* here
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and then substitute in the force exerted by
the wall in place of *F wall* here.
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So this is substituting equation 3 version b
in place of equation [1b]
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and then we'll call this result equation [1c]
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so we have *mg* times 0.35 meters divided
by 2*cos Θ* times 1.20 meters.
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Then we'll look at equation 2 version b
and substitute in what we have for
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the force on a foot
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equation [1c] in place of *F* in
equation 2 version b.
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So solving... rewriting equation 2 but
instead of the force on a foot,
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we are gonna write *mg* times 0.35 meters
divided by 2*cos Θ* times 1.20 meters
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and so we do that here.
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Now whole bunch of things cancel
which is convenient:
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the *mg*'s cancel, these 2's cancel and
*sin Θ* divided by *cos Θ*
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can be replaced with *tangent Θ*—
that's a trigonometric identity—
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and that's gonna equal 1.20 meters
divided by 0.35
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because the other thing we'll do here is
multiply both sides by 1.20
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and divide both sides by 0.35
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and *Θ* then is gonna be inverse tangent of
1.20 divided by 0.35
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which is 73.7 degrees.
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Then returning back to equation 2,
we'll call it version c,
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where we solve for the force on a foot
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that's going to be *mg*
divided by 2*sin Θ*.
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We have 500 kilograms—total mass of
the horse and rider—
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times 9.8 newtons per kilogram
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divided by 2 times *sin* of this angle
that we just found
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and that is 2550 newtons.
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So the force on a single foot is
2550 newtons,
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73.77 degrees above horizontal
towards the wall.
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Then the next question in part (b) asks
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what is the minimum coefficient of
static friction
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that's necessary to keep these
feet from slipping.
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So the total friction force is
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2 times the horizontal component of
a force on a foot
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and that also is going to be
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the coefficient of static friction
multiplied by
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the normal force which will in turn
equal the weight
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of the horse and rider.
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So we'll solve for *μ s* by dividing
both sides by *mg*
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so that's 2 times the force on a foot
times *cos Θ* divided by *mg*.
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So that's 2 times 2552.1 newtons
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times *cos* of 73.7398 degrees
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divided by 500 kilograms times
9.8 newtons per kilogram
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which is 0.292 and that is the minimum
coefficient of static friction.