# Video: Calculating the Torque Produced by a Force

A spanner is used by a mechanic to tighten a nut. The spanner is 20 cm long and to tighten the nut it must apply a torque of 12 N⋅m. What force must the mechanic apply to the end of the spanner that is opposite to the end that tightens the nut? The spanner turns through a horizontal circle to tighten the nut.

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### Video Transcript

A spanner is used by a mechanic to tighten a nut. The spanner is 20 centimeters long, and to tighten the nut, it must apply a torque of 12 newton meters. What force must the mechanic apply to the end of the spanner that is opposite to the end that tightens the nut? The spanner turns through a horizontal circle to tighten the nut.

Okay, so in this question, we’ve been told that we’ve got a nut and a mechanic is using a spanner to tighten that nut. So here’s our spanner, which we’ve been told it is 20 centimeters long. Now, in the question, we’ve been told that firstly the spanner happens to be 20 centimeters long. And secondly, we’ve been told that in order to turn the nut, we need to apply a torque of 12 newton meters.

Given this information, we’ve been asked to find the force, which we’ll call 𝐹, applied by the mechanic to the spanner. Specifically, this force is applied to the end of the spanner that is opposite to the one that tightens the nut. And that’s exactly how we’ve drawn it in the diagram. This is the end at which the force acts and this is where it ends up acting.

Now, the final sentence in the question actually tells us something quite interesting. We’ve been told that the spanner turns through a horizontal circle to tighten the nut. In other words, when the spanner turns, the end at which we apply the force follows a horizontal circle. Now, that’s a particularly badly drawn circle. But let’s imagine it is a circle for now.

So what is the relevance of this? Well, the relevance is not the fact that it’s a circle, but the fact that the circle is a horizontal circle. Because if the circle is a horizontal circle, then it means that the way that we’ve drawn the diagram is looking down on the nut and the force of gravity acts on the spanner and the nut into the screen. In other words, gravity doesn’t act this way or that way or this way or that way. It acts into the screen. And that’s important because gravity offers no help to the mechanic in exerting a force on the spanner. The entire force that is exerted is exerted by the mechanic. And that’s the whole point of the final sentence of the question.

So now that we’ve established that, let’s try and work out the force 𝐹. To do this, we can recall that a torque is defined as a force applied to an object multiplied by the perpendicular distance between the point at which the force is applied and the point at which this force acts. Now, as we’ve already seen the force is applied here and it acts here. So the distance between these two points is 20 centimeters, which is what we’ve been given already.

Furthermore, the distance between the point at which the force acts and the point at which the force is applied is measured in this direction. Now, this direction happens to be perpendicular to the direction of the force. Therefore, the 20-centimeter distance that we’ve been given in the question is the perpendicular distance. And we already know the torque that needs to be applied. It’s 12 newton meters, which means that we know the torque and the perpendicular distance and we can use this to work out the force 𝐹.

To do this, we rearrange the equation by dividing both sides of the equation by 𝑑. And we find that 𝑇 divided by 𝑑 is equal to 𝐹. Now, before we plug any values in, we need to convert everything we have to standard units. The torque already is in a standard unit of newton meters. However, the distance is in centimeters, which means that we need to convert this to meters. If we do so, then we’ll find the force in its standard unit of newtons.

So we can recall that 100 centimeters is equal to one meter. And if we divide both sides of the equation by five, we will find that 20 centimeters is equal to 0.20 meters. And at this point, we can plug in our values: 𝑇 is 12 newton meters and 𝑑 is 0.20 meters. Evaluating this fraction gives us a value of 𝐹. 𝐹 ends up being 60 newtons.

So we find that the mechanic must apply a force of 60 newtons to the end of the spanner.