Torque Due to Weight - Online Tutor, Practice Problems & Exam Prep

Torque is produced by an object's weight force, mg, acting at its center of gravity, which is typically the geometric center for objects with uniform mass distribution. For a fixed rod, the net torque can be calculated using the equation τ = Frsin(θ), where r is the distance from the axis of rotation and θ is the angle between the force and the lever arm. Understanding these concepts is crucial for solving physics problems involving rotational motion and equilibrium.

Torque Due to Weight

Hey, guys. So in this video, we're going to talk about an object's weight force, mg, producing a torque on itself. In other words, an object's own weight will produce a torque on that object. Let's check it out. So imagine you have a bar like this that is free, that can rotate about this point. So if it does spin in any way, it would be around here, kind of like your arm going like this. Right? Now there's a weight force that pulls in the middle like this, mg. Because if you have mass, you have weight, so there's something pulling down on you, which means you can think of it in terms of the weight of your arm is pulling down like this or pushing down like this. And because you're fixed here, it's trying to cause it to spin like this. Okay. It's trying to cause this. So knowing, so being aware of the torque and knowing how to calculate the torque due to weight, is really important because it's going to come up in a lot of questions. Alright? So I want to remind you, however, a few things real quick. So an object's weight, mg, always acts on the object's center of gravity, center of gravity. Now for most of you, the terms center of gravity and center of mass will be the same thing. So you can also think of this as center of mass. Okay? Most professors will actually not even draw a distinction between the two. Now, the way this works really briefly is that they are different only if your gravitational field is not the same everywhere. So there's, like, more gravity here than here. For most of you, you're not going to have to worry about that. And you can think that center of gravity really just means the same thing as center of mass. If those two things are different, then you have to look at your center of gravity, not your center of mass. But again, if your professor has made a distinction about that, you're used to that by now. If not, don't worry about it. If this sounds foreign, it's probably because he hasn't talked about this and you don't have to worry about it. Okay? So but the official answer, the correct answer is center of gravity. It's just that they're the same most of the time. If an object has what's called uniform mass distribution, uniform mass distribution means that the mass of an object is evenly distributed throughout. Okay? You can think of this as even mass distribution, very well distributed. It's the difference between a bar that has roughly the same mass everywhere or exactly the same amount of mass everywhere versus a bar that's much heavier, let's say on this piece right here. Right? Most of the time, you're going to have uniform mass distribution. And in this case, your center of gravity or center of mass, if they are the same, is going to be on the object's geometric center. What the heck is geometric center? It's just fancy for it's in the middle. Right? So if you have, let's say, this bar is 4 meters long, length equals 4 meters, it means that the mg is going to pull at a distance of 2 meters from the edges, right down the middle. Okay? Long story short, mg pulls in the middle. That's how it works for a vast majority of physics problems.

So let's look at this bar here. We have a bar, a cylindrical rod that has a mass m equals 20 and a length equals 4. And it has one of its ends fixed to an axis right here that is mounted on the floor. The rod makes 37 degrees above the horizontal, right there. And suppose you have a mass of 80, so that's your own personal mass, and you're going to stand on the other end of the rod. So you're up here with a mass, let's call it big M, of 80 kilograms. We want to know the net torque. Okay? So net torque, that's what we want to know. That is produced on the rod about its axis due to the two weight forces. Okay. So how much torque do you produce on the rod? About its axis, that's always how this works. This is just sort of extra language. Torques are always produced about an axis of rotation, due to the weight forces. So what we're doing is we're doing torque due to little mg plus the torque due to big Mg. And it says, you may assume the rod has uniform mass distribution. What that means is that the weight force on the torque on the rod will happen right at the middle. If this question didn't say that, you're going to assume that anyway. Cool? So unless the question says that the object doesn't have uniform mass distribution, you assume that it has uniform mass distribution. It's kind of like friction back in the day. If it didn't say anything about friction, you just assume no friction. Okay? It's fixed in place, so it does not move or rotate. So this object, the rod isn't going to move sideways or up and down. And actually, in this particular case, it doesn't even spin. Now just to be clear, it doesn't mean that there are no torques on it. It just means that for whatever reason, something is stuck here, it's made to not spin. Okay. You don't do anything with that. You still calculate the net torque on it. Alright. So we can calculate these two guys individually, torque of little mg, torque of big Mg, and then we're going to put them together. I'm going to leave a little space for the signs. And remember, torque torque is Frsinθ. In this case, the force gets replaced by mg because that is the force. So the torque of little mg is mgrsinθ. Now there are two forces and they are different. Right? This is the r for little mg, and this is the r for big Mg, and the angles are different as well. This is the angle for little mg, the angle for big Mg. Okay. So what we got to do is plug in these numbers and the sines. Okay. So remember the steps, draw our vector, figure out the θ, and then plug it into the equation. R vector is an arrow from the axis of rotation right here to the point where the force happens. So this is r for little mg. Okay? And the other mg is going to have an r that goes all the way to the end of the bar because you stand on the edge right here. This is our big Mg. Okay? So r little mg is half of the bar because mg acts in the middle. So the length of the bar is 4, so your r mg is 2, your r big Mg is 4. Okay. So we got the we got the r's figured out. What about the angles? What about the angles? So do we use this 37 here? And the answer is no. You do not use that 37. Be careful. That is the wrong answer. Let me draw it and show you why. So for this here, we have r from the axis to mg, and then this is your actual force which is mg. This is the angle that I need. That's the angle right here between these two, right, not this other angle here. Now notice that this forms a triangle. If this is a 37, it's a right triangle. This is a 90 so this has to be 90 minus 37. So this is 53. Very, very careful. We're going to use 53 instead. Okay? The same thing happens for the other mg. The only difference is that the arrow is longer but that doesn't matter. Our big Mg, mg is this way. This 37 still shows up right here but that's still not the angle we want. We want 53. Okay. Be very careful. The angle that you're being given is not the one that you're supposed to use. That happens quite a bit. Now finally, before we plug all this stuff in here, we got the angle, we got the r's. We can find the f's, just mass times gravity. Now let's look at the sine. Okay. So what we're going to do is pretend that this rod is your arm, right, and you're bei

Two kids play on a seesaw that has mass 20 kg, length 3 m, and its fulcrum at its mid-point. The seesaw is originally horizontal, when the kids sit at the edge of opposite ends (m,_LEFT = 25 kg, m,_RIGHT = 30 kg). Calculate the Net Torque from the 3 weights acting on the seesaw, immediately after the kids sit (simultaneously) on their respective places.

• A. 75 N•m clockwise
• B. 75 N•m counterclockwise
• C. 150 N•m clockwise
• D. 150 N•m counterclockwise

A guy standing straight up stretches out his arm horizontally while holding a 60 lb (27.2 kg) barbell. His arm is 64 cm long and weighs 45 N. Calculate the Net Torque that the barbell and the weight of his arm produce about his shoulder. You may assume that his arm has uniform mass distribution.

• A. −156 N•m
• B. −171 N•m
• C. −185 N•m
• D. −312 N•m

Here’s what students ask on this topic:

What is torque due to weight and how is it calculated?

Torque due to weight is the rotational force exerted by an object's weight acting at its center of gravity. It is calculated using the equation:

$\tau =Fr\sin \left(\theta \right)$

where $F$ is the force (weight, $\mathrm{mg}$), $r$ is the distance from the axis of rotation to the point where the force acts, and $\theta$ is the angle between the force and the lever arm. This concept is crucial for understanding rotational motion and equilibrium in physics.

How does the center of gravity affect torque due to weight?

The center of gravity is the point where the weight of an object acts. For objects with uniform mass distribution, the center of gravity is at the geometric center. The torque due to weight is calculated by considering the force acting at this point. If the center of gravity is not at the geometric center, the distance $r$ in the torque equation changes, affecting the torque value. Understanding the center of gravity is essential for accurately calculating torque in various physics problems.

What is the significance of the angle θ in calculating torque due to weight?

The angle $\theta$ in the torque equation $\tau =Fr\sin \left(\theta \right)$ is the angle between the force vector and the lever arm. It determines the effectiveness of the force in producing rotational motion. A force applied perpendicular to the lever arm ($\theta =90°$) produces maximum torque, while a force applied parallel to the lever arm ($\theta =0°$) produces no torque. Correctly identifying this angle is crucial for accurate torque calculations.

How do you calculate the net torque on a rod with uniform mass distribution?

To calculate the net torque on a rod with uniform mass distribution, you need to consider the torques produced by all forces acting on the rod. For a rod fixed at one end, the torque due to its own weight is calculated at its center of gravity (geometric center). If additional weights are applied, calculate their torques separately and sum them up. The net torque is the sum of individual torques:

$\tau =\tau 1_{}+\tau 2_{}+...+\tau n_{}$

Ensure to consider the direction (positive or negative) of each torque based on the rotational direction they cause.

What is the difference between center of gravity and center of mass?

The center of gravity is the point where the gravitational force acts on an object, while the center of mass is the point where the mass of an object is concentrated. For most practical purposes in a uniform gravitational field, these two points coincide. However, in a non-uniform gravitational field, they can differ. In physics problems involving torque due to weight, the center of gravity is typically used, especially when dealing with uniform mass distribution.