### Video Transcript

A sheet of mass 12 kilograms and
uniform density is balanced on a pivot, as shown in the diagram. What is the magnitude of the force
𝐹 that balances the sheet horizontally? Give your answer to the nearest
newton.

Okay, so in this question, we know
that we’ve got a sheet of mass 12 kilograms. And we know that the sheet has
uniform density. Now what the uniform density bit
tells us is that the mass of the sheet is balanced evenly along the length of the
sheet. And because of this, the center of
mass of the sheet is going to be at the center of the sheet. In other words, the center of mass
is going to be somewhere along here.

Now, we know that the sheet is 2.5
meters long. And because the center of mass of
the sheet is at the center of the sheet, we therefore also know the distance from
one end of the sheet to the center of mass. Now, this distance obviously is the
same as the distance from the center of the mass to the other end of the sheet. And this distance happens to be
1.25 meters.

Now, the reason that this is useful
is because we know that the weight of the sheet must act through its center of
mass. So we can add an additional force
onto the diagram, which is the weight of the sheet. We can calculate this weight by
recalling that the weight of an object 𝑊 is given by multiplying the mass of that
object 𝑚 by the gravitational field strength of the Earth 𝑔. We can also recall that 𝑔 is a
constant 9.8 meters per second squared.

Now, since we know the mass of the
sheet — we know that it’s 12 kilograms — we can work out the weight of the
sheet. So we say that the weight of the
sheet is equal to the mass multiplied by the gravitational field strength of the
Earth. And this weight ends up being 117.6
newtons. So we can replace the weight in the
diagram with 117.6 newtons.

Now what we’ve been asked to do in
this question is to find the magnitude or size of the force 𝐹 that balances the
sheet horizontally. And it’s important that the sheet
is balanced horizontally because this means that all of the clockwise turning forces
on the sheet balance out the counterclockwise turning forces on the sheet.

So what are the clockwise turning
forces on the sheet? Well, we have two forces that are
towards the right of the fulcrum — that’s this force and this force. Both of these are forces that are
trying to turn the sheet clockwise. And we can calculate the torque
generated by each one of these forces. Similarly, we can figure out the
counterclockwise turning forces.

Now, there’s only one force to the
left of the fulcrum and that’s force 𝐹. Force 𝐹 is the one that’s trying
to turn the sheet counterclockwise. And we can work out the torque or
turning force generated by force 𝐹 as well. Now, the torque or turning force is
defined as the force exerted on an object multiplied by the perpendicular distance
between the point at which the force acts and the fulcrum.

So let’s start by focusing on the
clockwise torques. We can say that the total clockwise
torque on the sheet, which we’ll 𝑇 sub clockwise, is equal to the torque generated
by the weight plus the torque generated by the 25-newton force. So first of all, the torque
generated by the weight, well, the torque generated by the weight is equal to the
weight itself — that’s the force which is 117.6 Newtons — multiplied by the
perpendicular distance between where the force is applied and the fulcrum.

So the distance in question is the
distance between the point at which the force is applied — that’s here — and the
fulcrum — that’s here. So we need to figure out what this
distance is. Now that distance is simply this
distance which we know is 1.25 meters because the center of mass is halfway long the
length of the sheet. And from this, we subtract this
distance, which we know to be 0.65 meters. So the distance which we’ll call
𝑥, we can say that 𝑥 is equal to 1.25 meters minus 0.65 meters. And so we find that 𝑥 is equal to
0.60 meters.

And then coming back to the torque
due to the weight, we multiply the force which is the weight by the perpendicular
distance which is 0.60 meters. But this is not the only clockwise
torque on the sheet. We need to consider the clockwise
torque due to this force. And so we add on the force which is
25 newtons multiplied by the distance between the fulcrum and the force. Now that distance has been given to
us already as 0.95 meters because that’s the distance between this point and this
point which is where the force 𝑥.

So coming back to torque, we
multiply the force by the distance which is 0.95 meters. And at this point, we’ve
reconsidered all of the forces trying to turn the sheet clockwise. So let’s look at an expression for
the anticlockwise turning force. Let’s now move on to considering
the counterclockwise turning force.

Let’s call this 𝑇 sub
counterclockwise. And let’s now remember that the
only force that’s acting towards the left-hand side of the fulcrum is the force
𝐹. This is the only force that’s
trying to turn the sheet counterclockwise. So we need to work out the torque
generated by this force. Well, the torque generated by that
force is simply the force itself which is called 𝐹 multiplied by the perpendicular
distance to the fulcrum — that’s 0.65 meters — at which point we have an expression
for the counterclockwise torque.

Now, if this sheet is to be
balanced on the fulcrum, then the clockwise torques and the counterclockwise torques
have to cancel each other out. In other words, the magnitudes or
the sizes of the clockwise torque and the counterclockwise torque must be equal. And so we can put a big equal sign
between them. The clockwise torque is equal to
the counterclockwise torque.

Let’s then get rid of the 𝑇 sub
clockwise and the 𝑇 sub counterclockwise. At this point, we know every value
in the equation, apart from 𝐹. So let’s rearrange to find 𝐹. To do this, we can divide both
sides of the equation by 0.65, at which point does 0.65 on the right-hand side
cancel leaving us with only 𝐹 on the right-hand side. We can then plug the value on the
left-hand side of the equation into our calculators to give us a value of 𝐹 to be
145.092 dot dot dot newtons.

Now, remember we need to give our
answer to the nearest newton. So we need to round this answer to
the nearest newton, which means we need to round to this value here. Now, it’s this next value here —
the one after the decimal point — that will tell us what happens to the value in the
ones column. Now since this value is zero and
zero is less than five, this value will stay the same. It’s is not going to round up. And so to the nearest newton, the
value of 𝐹 is 145 newtons. And this is our final answer to the
question.