A 2-m long bar is free to rotate about an axis located 0.7 m from one of its ends. Two forces act on the bar, F 1 = 100 N and F 2 = 200 N, and both make 30° with the bar. Find the Net Torque on the bar. Use +/− to indicate direction.

Hey, guys. Hope you try this practice problem. Here we have 2 forces producing 2 torques on a bar. Let's check it out. So we have a 2 meter long bar that is free to rotate about an axis located 0.7 from one of its ends. So this is the axis around which it's, free to to rotate. The bar is 2 meters long. The axis is 0.7. So from this, you are able to determine that this is the 0.7 distance because it's the shorter of the 2. And this is the other distance, which must be if the whole thing is 2, this is a point 7, this must be 1.3 meters away. We got the forces f one equals 100 and f two equals 200, and they both make an angle of 30 degrees with the bar. So 30 degrees here and 30 degrees here. We wanna find that net torque and we wanna use signs to indicate direction, clockwise or counterclockwise. So the net torque, torque net, is simply torque 1 +2. We're gonna calculate these 2 guys individually and put it all together. So torque 1 is f1r1sine of theta 1. I put some space here for us to determine whether it's going to be positive or negative, which we can do that at the end. So f 1 is a 100. The r one, let's draw our r vector. R, remember, is a vector from the axis to the force. So force 1 is right here. So r one is this. That's my r1. R2 is this. That's my r2. Okay? So the length of r one is simply point 7. So I'm going to put a point 7 right here. Sine of theta 1. Let's make space for theta 1 and figure that out. So what's the angle between theta one is the angle between r one and f one. What is that angle? So r one is this way. F one is this way. Remember, we can use this angle here. That is the angle between those two arrows, those two lines. So that angle is good to go. We don't have to move anything around. Okay? So if you do all of this, you get 35. Now is that 35 positive or negative? Again, let's look at this piece of the bar. We don't have to worry about the whole bar. So look at it this way. I have my arm here and I have a force that's pulling it down this way. So this is the axis of rotation. You're trying to do this to the bar, right? So so you're trying to do this to the bar, you have to be careful in this part. You're trying to get the bar to spin this way with torque 1. This is in the direction of the unit circle. It's opposite to the clock, direction of the unit circle is positive. Okay? So this guy is going to be positive 35. What about torque 2? Torque 2 is f 2 r2 sine of theta 2. Since we're in the topic of direction, we just talked about that, let's talk about the direction for torque 2. So we look at this piece from the axis to the point where the force happens. Imagine this is the door now this way. Here's the axis right here. Right? And I'm pulling like this, so I'm trying to I'm trying to cause this thing to spin in this direction, right. So it's this, It's in this direction here, which is in direction of the unit circle, so it's positive as well. In this example, they're both positive. That's nice, which means that they're not trying to cancel each other out. They're actually helping each other out, and they're both spinning the same way. So I have positive, the force of f 2 is 200. The the distance here is 1.7 I'm sorry, 1.3. Right? So the r vectors from here to here. The the length of r two is 1.3. And then the angle we're supposed to use is the 30 because here's our 2 and here's f 2. This is the angle between those two vectors. So this guy is good to go, 30 degrees. Okay? If you multiply all of this, you get a 130, and these are both newton meters. So if you combine them, torque net is simply the addition of those 2 guys, plus 35, plus plus 130. So this is positive 165 Newton meter. Okay? Trickiest thing here was to figure out direction. The angles were easy. So that's it for this one. Hopefully, it makes sense. Let me know if you have any questions and let's keep going.

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14. Torque & Rotational Dynamics

Net Torque & Sign of Torque

Physics

14. Torque & Rotational Dynamics

Net Torque & Sign of Torque

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Multiple Choice

A 2-m long bar is free to rotate about an axis located 0.7 m from one of its ends. Two forces act on the bar, F₁ = 100 N and F₂ = 200 N, and both make 30° with the bar. Find the Net Torque on the bar. Use +/− to indicate direction.

−95 N•m

+95 N•m

−165 N•m

+165 N•m

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Verified step by step guidance

Identify the pivot point, which is located 0.7 m from one end of the bar. This is where the bar rotates.

Calculate the perpendicular distance from the pivot to the line of action of each force. For F1, the distance is 0.7 m, and for F2, the distance is 1.3 m (2 m - 0.7 m).

Determine the component of each force that contributes to the torque. Since both forces make a 30° angle with the bar, use the sine function to find the perpendicular component: F1_perpendicular = F1 * sin(30°) and F2_perpendicular = F2 * sin(30°).

Calculate the torque produced by each force using the formula τ = r * F_perpendicular, where r is the distance from the pivot. For F1, τ1 = 0.7 m * F1_perpendicular, and for F2, τ2 = 1.3 m * F2_perpendicular.

Add the torques considering their direction. If F1 causes clockwise rotation, it is negative, and if F2 causes counterclockwise rotation, it is positive. The net torque is the sum of τ1 and τ2.