# Torque Example

by Jared Rovny

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00:01 Here's an example of how to solve a torque problem and it supposing that you have a 10 kg shop sign and it's hanging from the center of a pole that has a pivot to it.

00:11 It's held up by a 2 meter long rope which is connected to the end of the pole.

00:15 And it's also connected to the wall 1 meter above the pivot point.

00:20 We could ask ourselves, how much tension would be on this rope? Which is practically a very important question because we want to buy a rope strong enough to hold up our sign.

00:28 So give this problem a try and see if you can use the torque equations and the idea of the lever arm as we've just discussed to solve a torque problem like this one.

00:37 So if you've given this problem a try, hopefully your answer looks something like what I'm about to show here.

00:43 We see in this problem is we have a force which is the force of gravity acting on our sign directly downwards.

00:51 So we know that this force of gravity will be equal to the mass of our sign times the gravitation acceleration.

00:58 We also have a tension being applied by this rope.

01:02 So we'll just call this F tension for example.

01:05 We have some angle, theta, which we will just define here as we move.

01:09 We have our pivot point because this is point around which everything this problem will be rotating.

01:14 And then in terms of numbers we're told that this distance is 1 meter from the pivot point.

01:21 And then also that we have a 2 meter long rope so we know that this is 2 meters long in this direction.

01:30 So given this fact what we would like to do is try to find a way to put the sign in equilibrium.

01:35 We don't want the sign to be moving.

01:37 It's not rotating down or up, it just sitting exactly where it is, so what we would do is do exactly what we did with Newton's second law when nothing is moving when there is no acceleration and that is to write a torque equation for all the torques in the problem and then say that the total torque acting in the system is equal to zero.

01:54 So let's figure out what all the torques are acting on our bar here.

01:58 So the torque acting on this bar if we consider this pivot point here, is simply the gravitational force here acting directly downwards and then the tension force here which is acting in a strange angle theta.

02:11 And so let's write those torques.

02:13 The torque in total is equal to, now what we need again is to write the torque as a force times a distance so let's first look at this force right here.

02:23 The distance is something we don't know.

02:25 This is some distance here. Let's just call the length of our bar, so that we can at least talk about this in our equations.

02:31 Let's call it R. We have some bar of length R and it's very important to notice when you're going through your problem like this, and you are not given a particular variables so for example here were not given the length R.

02:44 You can still introduce a variable on your own.

02:47 You just hope that by the end, the variable disappears because you don't know what it is.

02:51 But we're just going to follow this through and see what happen if we simplify our equations.

02:55 So if we call this distance R, and the distance at which the gravitational forces is acting is R over 2 and so the torque from the gravitational force is force which is the gravitational force times distance which is R over 2.

03:12 Our few more things to consider with this torque though.

03:15 First of all, what is the sine of theta? We have that the radius from the pivot point is directly off to the right.

03:22 And then we have a force from our sign which acting directly downwards.

03:25 And so the angle between these two, the R and the F and our torque equation is 90 degrees and so we actually do not need to worry about the sine of theta because this torque the force rather acting from the sign is exactly perpendicular to the distance from the pivot point, so nothing to worry about there.

03:41 The other thing we have to consider with this force, though, is that it's acting in a clockwise direction.

03:46 Notice that relative to my pivot point if my sign is pulling downward the entire bar is going to want to rotate clockwise which means that in our sign convention for torques trying to rotate things to the left or to the right, this would be consider a negative torque because again it's trying to twist the system in a clockwise direction.

04:04 So we'll make sure we include our minus sign here.

04:06 So we have a minus and here's our torque from the sign.

04:11 So now we have to just consider the torque acting from the rope from the force of tension.

04:16 And this is one, that's a little bit trickier.

04:21 One, is that we have our force acting at the distance R and then we have sine of theta, so just writing it out very naively almost just exactly in this many letters.

04:33 We have the force times the distance which we have introduced here, times the sine of the angle theta whatever theta angle is.

04:41 So one way to solve this we just be to look at the geometry of your problem and immediately know what the sine of theta is because the sine of theta is always define in terms of some triangle.

04:50 If you look at this problem what we have is a triangle with a one meter long leg and then a 2 meter long hypotenuse if you consider the entire triangle form by the bar the wall and the rope.

05:01 So if you look at that triangle what you see, is that again, the opposite side is 1 meter in length.

05:06 The hypotenuse is 2 meter in length and since the sine of theta by definition is the opposite side over the hypotenuse, we have that the sine of theta, is equal to one half.

05:18 So this is one easy way to figure it out what the torque would need to be.

05:21 It just to say that the torque is F times R times one half from your sine of theta.

05:28 Another way, if you don't want to rely on that sort of trigonometric argument is to again consider the lever arm for this problem.

05:34 So let's write where the lever arm would be.

05:38 And again we draw a line directly from our pivot point to the line of force where the force is the force of tension acting upon this direction and so we would extend this line along the rope and this would be our line directly to the line of force and this length here would be our lever arm.

05:53 And now we again have to come up with the trigonometric argument.

05:57 We have a length, a lever arm here and we also have a distance R and so this lever arm would also be R times the sine of theta.

06:12 So now that we have an expression for the torque let's try to simplify it and again we're going to consider that the entire torque acting on our object is zero because the entire thing is not actually rotating one way or the other.

06:25 So if we say that zero is our torque we have minus FG times R over 2 plus force of tension times R times the sine of theta, where again we've seen that this is simply one half.

06:42 Simplifying this we have FG times R over two is equal to the force of tension times R times one half.

06:51 This R over 2 fortunately are cancels from both side.

06:55 We can divide both sides by the radius R over distance the length of our sign of the pole.

07:00 And its entirety divided by 2 we can divide both side by that whole quantity and it goes away which is exactly what we were hoping for when we introduced the variable R that it wouldn't come back into this problem because we don't actually know what that entire length is.

07:13 So dividing this R over two from both side we simply have the tension force must be equal to the gravitational force.

07:21 And so the tension force is equal to the mass of the sign times the gravitational acceleration which in this case it simply equal to 10 kg, times the gravitational acceleration which we can just call 9.8 here, be very accurate and this is because we don't anymore work to do in this problem.

07:39 So it doesn't cost this anytime to do this.

07:42 And we can say that the tension in the rope is 98 Newton's.

07:47 Again, equal to the weight of the sign M times G.

The lecture Torque Example by Jared Rovny is from the course Equilibrium.

### Included Quiz Questions

1. 0.5 m
2. 2 m
3. 4 m
4. √3/2 m
5. 4.5 m
1. τ = 1 N.m
2. τ = 3 N.m
3. τ = 2.5 N.m
4. τ = 1.5 N.m
5. τ = 2 N.m
1. 5 N
2. 10 N
3. Depends on the angle which the forces make with the rod
4. 2.5 N
5. 15 N

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