WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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In this problem, we're told to pretend that
the chain there doesn't exist or it's broken,
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in which case the horizontal force
on the foot of one of the sides
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of the sandwich board has
to be provided by friction.
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We're going to figure out what is the
coefficient of static friction that's needed,
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the minimum anyway that's
needed to make a force
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that's big enough to prevent
this foot from slipping.
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So we're going to consider just
one half of the sandwich board.
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It's always good to take
a complicated problem
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and try to reduce it into a smaller
problem that's easier to think about.
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We have a bunch of forces on here.
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We have the normal force straight
up provided by the ground,
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we have the friction straight to the left,
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we have gravity acting
at the center of mass
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which is the geometric center of this board
since we're told that it's a uniform board
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and the mass is distributed evenly
throughout the whole length of this board.
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So its center of gravity is
in the physical center.
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Then we have the hinge force which
I've drawn straight to the right
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and I need to explain why we
know that it's straight to the right.
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Because being a hinge it could conceivably
have a component that is vertical
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but we'll see that it is not in this case
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because we'll turn our
attention to this picture here.
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Consider this whole sandwich board
as being a single point mass
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and the total force downwards
on it would be
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two times the force of gravity on each side.
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So I have *F g* here representing the force
of gravity on one side of the board,
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and so if you reduce this whole
sandwich board to a dot,
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the total force down on it
would be two times *F g*.
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And order for it to not
accelerate vertically,
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the ground will have to provide a force
upwards that's of the same magnitude.
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So two times this normal
force in other words,
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which is the only force upwards on
the whole thing is going to be --
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well, it works out to saying that
the normal force upwards
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has to equal the gravity downwards.
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Since that's the case, there can't
be any upwards hinge force
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because this normal force is
accounting for the only force
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that is upwards on the
sandwich board as a whole.
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Do I need to say anymore about that?
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You know, if the sandwich
board was not symmetrical
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and maybe one side was long
and the other side was short,
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then things would get more complicated
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and there would be a vertical component
to the hinge force for sure.
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We'd have to be given more
information to solve it
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because you'd no longer be able to say
that two times *F n* is the force upwards
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because there'd be force
normal on the right side
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and there'd be force normal on the
left side and they would not be equal.
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That means you'd have *F nl* plus
*F nr* here instead of two *F n*
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but then you'd have two unknowns.
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So that would complicate things.
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But fortunately we have
a symmetrical system
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in which case the normal force
on each foot upwards is the same
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and so that's why we're able to say two *F n*.
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Yeah, so that's how we conclude
that the normal force upwards
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equals gravity downwards.
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Since that's the case there is no place
for having a vertical force on the hinge.
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Okay.
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So, well I guess I could expand
on that slightly more.
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Let's suppose you still were to insist that
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there is a vertical component
of the hinge force.
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You'd have to say that in the y
direction the hinge force vertical
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plus the normal force would
equal the gravity downwards
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and then you could solve
for the hinge force
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and say that it's gravity minus normal force.
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But since we've just said that gravity
and the normal force are equal,
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that makes this zero.
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So there is the more rigorous
reason for saying
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vertical component of hinge force is zero.
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Okay, there.
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Now, with that out of the way,
we can think about this picture
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and consider our three equations which we
get from the two conditions for equilibrium,
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one being that the translational
forces have to add up to zero
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in both directions, x and y,
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and then we'll say that the total torque
has to add up to zero as well.
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So to say that the horizontal forces
add up to zero is the same
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as saying the forces to the right
equal the forces to the left.
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So the hinge force to the right
equals the friction force to the left.
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I substituted coefficient
of static friction
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multiplied by normal force
in place of the friction force.
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Then vertically speaking we have
normal force equals gravity,
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we've kind of already said
that but just saying it again,
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and that is *m g*, mass being the mass
of one side of the sandwich board.
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It's eight kilograms divided by
two which is four kilograms.
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Then the total torque has to equal zero and
I've taken the pivot to be at the ground
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but you can take it at the hinge if you
like, doesn't really matter where.
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I chose to choose the pivot at the ground
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because then we have only
two torques to consider.
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We have this due to gravity and
the torque due to the hinge.
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If we had chosen the hinge to be the pivot,
then we'd have the torque due to gravity,
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the torque due to friction and the
torque due to the normal force
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and we'd have three terms to consider.
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So I think it's a little bit
simpler to have only two.
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So we're taking the pivot to be at the
point of contact with the ground.
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Okay.
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So we have the hinge force multiplied by
its perpendicular distance to the pivot
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which is this 1.3 meters height
of the sandwich board.
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That's going to be the torque
which is clockwise,
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and that's going to equal
in magnitude the torque
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going counter-clockwise
due to the force of gravity
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which will have a perpendicular
distance to the pivot of 1.1 meters
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which is the total width of the
sandwich board divided by four.
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So 1.1 divided by two would
take you to the hinge
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and then this center of gravity is halfway
between the center point and the foot
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and so we divided by four
to get 0.275 meters
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as the lever arm perpendicular for gravity.
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So we've equated those two
things, those two torques,
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force of gravity is *m g* and then
we're dividing both sides by *r h*
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the lever arm for the hinge to
solve for the force of the hinge.
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So that's four kilograms times 9.8 newtons
per kilogram, times 0.275 meters,
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lever arm for gravity, divided by
1.3 meters, lever arm for the hinge.
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That gives 8.29 newtons
and that is horizontal.
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That's the answer for part B.
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The answer for part A, the
coefficient of static friction,
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we get by rearranging this formula here
and we divide both sides by *F n*
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to get that coefficient of static friction
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is the hinge force divided
by the normal force.
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So that's 8.2923 newtons
divided by four kilograms
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times 9.8 newtons per kilogram,
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giving a coefficient of
static friction of 0.211.