# Homework Help: Equilibrium Question involving Torque

1. Jul 12, 2014

### mlb2358

1. The problem statement, all variables and given/known data
The crane shown below has a mass of 4000 kg and a base of 3.4 m. The arm of the crane is 22 m and attaches to the center of the crane. If the arm is placed at an angle of 30°, what is the largest mass that the crane can hold off the ground without tipping?

2. Relevant equations
∑τ = 0 when there is rotational equilibrium.

3. The attempt at a solution
The answer to the question chooses the pivot as the right edge of the body of the crane. When I do this, I get the correct answer (400 kg), but at first I chose the pivot to be at the center of the arm of the crane. When I do this, I get 11cos(30)mg = 11cos(30)(4000)g, which makes m = 4000 kg. I don't know why this method is giving me the wrong answer. I always thought that for rotational equilibrium net torque would equal zero at any point in space. Why should the answer vary with what I choose as the pivot? I guess I could also be omitting a force in calculating the torque, but I don't think I am.

2. Jul 12, 2014

### Orodruin

Staff Emeritus
You are correct, the net torque should be zero regardless of the point chosen. How did you describe the force of the ground on the crane?

3. Jul 12, 2014

### mlb2358

Thanks for the reply. I guess the ground would exert a normal force on the crane, but wouldn't this simply be equal and opposite to the weight of the crane? If not, then I'm really not sure how to incorporate it.

4. Jul 12, 2014

### Orodruin

Staff Emeritus
Yes, equal and opposite: but where is its point of application?

Edit:
Also, for equilibrium when he weight is lifted, it is the weight of the crane plus the weight of the load.
What is the torque of the crane itself around its center of mass?

5. Jul 12, 2014

### mlb2358

I see now. The ground exerts a normal force at the right end of the body of the crane. The normal force can be found by finding the net force in the y direction and equating it to 0. This gives the correct answer. As for the torque exerted by the crane arm, no separate information is given for it. I think this means we should assume that it has been taken into account when the center of gravity was calculated, or we are to assume its massless. Thanks for your help!

6. Jul 12, 2014

### Orodruin

Staff Emeritus
Yes, but to be clear, this is just in the limiting case that you are after. Without any load, it would be right below the center of mass and as the load increases, the point of application would move to the right. As it is not possible to apply a force to the right of the crane, the crane tips when equilibrium would require this. The limiting case is therefore given by the situation you describe with the application at the right end. This is also why you can ignore this force when using that point for computing the equilibrium.l

7. Jul 12, 2014

### mlb2358

Thanks, I didn't realize that until you mentioned it. Also, just to be clear, the point of application moves to the right so that the length of the lever arm decreases and thus the magnitude of the torque produced by the normal force can be minimized. I guess its also interesting to note that as the mass gets larger, the magnitude of torque created by the normal force increases (if we disregard the change in the length of the lever arm) and so does the magnitude of the torque created by the hanging mass. This means that to keep equilibrium conditions, the lever arm of the torque created by the normal force must decrease in magnitude by enough to compensate for both of these changes. Just want to make sure I understand things clearly. Thanks again!

8. Jul 12, 2014

### dean barry

If the distance from the tip of the boom to the centre of the crane is 22 m, then the horizontal distance of the load = 17.35 m from the pivot point, so the max load = ( 4000 * 1.7 ) / 17.35 = 391.93 kg