# Question Video: Torque Physics • 9th Grade

A sheet of mass 12 kg 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.

06:07

### 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.