Question Video: Relating the Speed of a Conductor to an Induced Emf | Nagwa Question Video: Relating the Speed of a Conductor to an Induced Emf | Nagwa

# Question Video: Relating the Speed of a Conductor to an Induced Emf Physics

The forces acting on an object are shown in the diagram. Find the magnitude of 𝐹. Answer to two decimal places. Find the magnitude of the reaction force. Answer to two decimal places.

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### Video Transcript

The forces acting on an object are shown in the diagram. Find the magnitude of 𝐹. Answer to two decimal places. Find the magnitude of the reaction force. Answer to two decimal places.

Alright, so in this question, we’ve been told that we’ve got an object, and we’ve been shown the forces acting on this object in the diagram. The first thing that we’ve been asked to do is to find the magnitude or size of the force 𝐹. Now in the diagram, this is the force 𝐹. It’s acting on the right-hand end of the object shown in the diagram.

Now as well as finding 𝐹, we’ve been also asked to find the magnitude or size of the reaction force. This is the reaction force. It’s acting in an upward direction on the object at the fulcrum. And it’s caused by all of the other forces, specifically the 10-newton force, the 15-newton force, and the force 𝐹 pulling down the object into the fulcrum.

Hence, the reaction force is also the contact force between the object and the fulcrum. And so, because these three forces are pulling the object down into the fulcrum, the fulcrum acts with an equal and opposite force according to Newton’s third law of motion, and it pushes upwards against the object. And that’s the reaction force. We’ll come back to that in a second though.

Let’s first look at finding out the magnitude of 𝐹. To do this, we first need to realize that out of all the downward acting forces on the object, there are two forces that are acting towards the right of the fulcrum; that’s the 15-newton force and the force 𝐹. And there’s one force that’s acting to the left of the fulcrum; that’s the 10-newton force.

Now the two forces to the right of the fulcrum, the 15-newton force and the force 𝐹, are trying to turn the object, this blue plank here, clockwise. And the 10-newton Force to the left of the fulcrum is trying to turn the same object counterclockwise. Now in the diagram, the object is shown to be balanced.

Therefore, the forces trying to turn the object clockwise and counterclockwise must also be balanced. Because if this wasn’t the case, the object would indeed turn either clockwise or anticlockwise, depending on whichever forces resulted in a larger torque.

And in fact it is torque that we need to study here because, remember, when a force is applied to an object away from the fulcrum — and just like we have here, the forces acting in a perpendicular direction to the distance between the point at which the fulcrum is and the point at which the force acts — there’s going to be a resultant torque.

Now, torque is defined as the force applied multiplied by the perpendicular distance to the force, in other words, distance we discussed earlier, the distance between the fulcrum and the point at which the force acts. So we can work out the total clockwise torque on the object and the total counterclockwise torque on the object as well.

Once we’ve done this, we need to equate the clockwise torque with the counterclockwise torque. In other words, we say that the clockwise torque, which we’ll call 𝑇 sub clockwise, is equal to the counterclockwise torque, which we’ll call 𝑇 sub counterclockwise. This is because, as we discussed earlier, the torques must be equal in order for the object to be balanced on the fulcrum.

So first of all, let’s actually go about working out the values of 𝑇 sub clockwise and 𝑇 sub counterclockwise. Let’s start with 𝑇 sub clockwise. 𝑇 sub clockwise is equal to the torque exerted on the object due to this force plus the torque exerted on the object due to this force.

In other words, 𝑇 sub clockwise is equal to — let’s start with this force first. What the torque exerted by that force is equal to 15 newtons, which is the force multiplied by the distance from the fulcrum. Now this distance happens to be 15 centimeters as we’re told in the diagram. So we multiply our 15-newton force by the 15 centimeters.

And then to this expression, we need to add the torque exerted by the second force 𝐹. So to this expression, we add the force 𝐹 multiplied by the distance between the point at which the force acts and the fulcrum. Now we’re told in the diagram that this distance is 75 centimeters. So we multiply the force 𝐹 by 75. And at this point, we realize that there are no other forces trying to turn the object clockwise.

Therefore, this expression here on the right is going to give us the clockwise torque on the object. So let’s work out the counterclockwise torque, 𝑇 sub counterclockwise. Well this torque is equal to the force that’s trying to turn the object counterclockwise, which is 10 newtons, multiplied by the distance between where the force is acting and the fulcrum.

Now this distance we’ve been told in the question is 25 centimeters. And so, we multiply the 10- newton force by the 25 centimeters. Now at this point, we have expressions for both the clockwise torque and the counterclockwise torque. So we can substitute them both in to this equation.

Giving ourselves more room to work on the right-hand side of the diagram, we can say that the clockwise torque, which is equal to 15 newtons multiplied by 15 centimeters plus 𝐹 multiplied by 75 centimeters, is equal to the counterclockwise torque, which is 10 newtons multiplied by 25 centimeters.

Now at this point, the only thing that we don’t know in this equation is the force 𝐹. So we can rearrange our equation to find out the value of 𝐹. To do this, let’s first subtract the value 15 times 15 from both sides of the equation. This way the 15 times 15 on the left-hand side cancels with the negative 15 times 15. And so, what we’re left with overall is just 𝐹 times 75 on the left-hand side.

And on the right, we’ve got 10 times 25, which we had already, minus 15 times 15. Now at this point, we can divide both sides of the equation by 75. This way, we’ll be isolating 𝐹 on the left-hand side, because on the left-hand side what we have is 75 divided by 75, which just leaves us with 𝐹.

And so when we neaten everything up, what we’re left with is 𝐹 is equal to 10 times 25 minus 15 times 15 divided by 75. Now at this point, all we have to do is to evaluate the fraction on the right-hand side. Well first of all, just evaluating the numerator gives us a value of 25. And so, what we’re left with is 25 divided by 75.

Now this is equal to one-third or 0.3 recurring. However, what we’ve been asked to do is to give our answer to two decimal places. So we need to use this value here. And to two decimal places — one, two — our answer is 0.33 newtons because, remember, we’re calculating the force 𝐹 and it’s going to be in newtons. And so our final answer to the first part of the question is 0.33 newtons. That’s the value of 𝐹.

Now for the second part of the question, we’ve been asked to find the magnitude or size of the reaction force. That’s this force labelled here. So let’s go about doing that by first giving ourselves again a little bit more space to work with. There! We’ve given ourselves a little space to work with.

No! Of course we meant space on the screen. Anyway, so let’s go about finding out the magnitude of the reaction force. So to work this out, what we actually need to do is to consider the three forces that are acting in the downward direction on the object altogether, because think about it this way: all of these forces are pulling the object down into the fulcrum, and because the object is being pulled into the fulcrum, the fulcrum is going to push back against the object with the reaction force.

Specifically, as we mentioned earlier, what we need to invoke is Newton’s third law of motion. Newton’s third law actually tells us that if an object called object one exerts a force on another object, object two, then object two exerts an equal and opposite force on object one. So in this case, we’ve got the object, which we’ll call object one, and we’ve got the fulcrum, which is object two.

Now the three downward forces are pulling the object, object one, into the fulcrum. So this object is exerting a force of 10 newtons plus 15 newtons plus 𝐹 onto the fulcrum. Hence, the fulcrum is going to exert a force of 10 newtons plus 15 newtons plus 𝐹 in the opposite direction onto the object. And specifically, that force is the reaction force. So the reaction force has a magnitude or size of 10 newtons plus 15 newtons plus the force 𝐹.

And as we’ve seen already, it acts in the opposite direction to the three forces. It acts in the upward direction. However, the question only asks us to worry about the magnitude, so we don’t need to worry about directions. And as we’ve seen already from the first part of the question, the value of 𝐹 is 0.33 newtons, at which point all that’s left for us to do is to just add these three values up on the right-hand side.

And this gives us the reaction force to be 25.33 newtons. Now, we’ve been asked to give our answer to two decimal places, but we already have our answer to two decimal places: first decimal place and second decimal place. And so we have our final answer: to the first part of the question, the magnitude of the force 𝐹 is 0.33 newtons; and for the second part of the question, the magnitude of the reaction force is 25.33 newtons.

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