# Homework Help: Torque and moments

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1. Oct 12, 2016

1. The problem statement, all variables and given/known data

Regarding part (c) (i), as we know that we can consider any of the convenient point as our pivot. If I consider the point where the 5kg mass (center of mass of the bag) acts, the asnwer turns out to be wrong. The correct answer has 4kg mass as pivot
2. Relevant equations

14(0.5x)=m(0.5x)+4(0.5x)
m=mass to be found
x=distance from the 5kg mass I have considered as pivot
3. The attempt at a solution
According to the equation above, x cancels out and we should get m=10kg, but the answer is 4+5(1/2)=16.5kg. Kindly explain why my approach is wrong and why can't we take the center of mass of the bag as our pivot?

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2. Oct 12, 2016

### Staff: Mentor

You can take the center as your axis, but you must include the torque due to all forces. There's an upward normal force acting at point C. (That's one reason to take point C as the pivot.)

3. Oct 12, 2016

Which upward normal reaction force is acting at point C. I can't figure out

4. Oct 12, 2016

### Staff: Mentor

You'll need to apply another equilibrium condition to solve for it. (That's why C is a good choice for the axis: The normal force doesn't exert a torque about that point and thus you don't care what it is.)

5. Oct 12, 2016

Here I just got to know that a normal force exists at point C. In any other scenario, I may repeat the same mistake of taking the wrong pivot. Therefore, kindly spare some of your precious time in answering what force it is? I'd be patiently waiting for your response.

Thanking you in anticipation :)

6. Oct 12, 2016

### Staff: Mentor

I'm not sure I understand your question. You do realize that the ground is supporting the suitcase with an upward force at C, right? To figure out what that force is, you'll need a second equation.

7. Oct 12, 2016

You got my question right, but the reaction force will only exist if there's a contact with the ground. Here, the bag is being lifted up the ground and there's no apparent contact

8. Oct 12, 2016

### Staff: Mentor

The bag is still in contact with the ground at point C. (It's just being barely lifted so that end of the bag at point D just leaves the ground.)

If the bag was not supported by the ground, there would be no equilibrium.

9. Oct 12, 2016

Ahh! I see
But again, won't the normal reaction be cancelled out by the weight?

10. Oct 12, 2016

### Staff: Mentor

No. That would be true if the bag were just sitting on the ground. But someone's pulling it up. You'll need to figure out that force it you still want to use the center as your axis. (Which is fine! You need to prove to yourself that it still works.)

11. Oct 12, 2016

I do concede that my method cannot be applied. The point is: at C, although the bag isn't sitting on the ground, a part of weight above it is acting, and it should provide an equal reaction according to the part of that weight and make the moment zero. No?

12. Oct 12, 2016

### Staff: Mentor

I don't think that's a helpful way of looking at it. Better to just say: Since it's in equilibrium, the net force must be zero. Since there's an upward force on the handle, the normal force from the ground does not need to equal the full weight. (And the bag is in contact with the ground at point C.)

13. Oct 12, 2016

I am still not clear that how does the normal reaction causes a moment about the axis. Nonetheless, thank you for giving your precious time and helping me out.

14. Oct 12, 2016

### Staff: Mentor

The normal force is like any other force acting on the bag.

15. Oct 13, 2016

### CWatters

Draw the free body diagram. It would show the normal force acting at point C. That creates a torque about any point except point C.

16. Oct 13, 2016

But weight will cancel that force out. No?

17. Oct 13, 2016

### Staff: Mentor

No.

18. Oct 13, 2016

### CWatters

No, and in any case what do you mean by "cancel out"?

I probably shouldn't draw it for you but here is a partial FBD anyway. I have left off the magnitude of the forces and assumed the case is empty. Got to leave you some work to do :-)

Note that the weight acts at the centre of mass (not at the wheel if that's what you were thinking). The Normal force acts at the wheel and there is another force on the handle. I have assumed the force at the handle acts vertically but it need not.

19. Oct 13, 2016

### CWatters

Just to clarify this.

Technically the weight of the case is distributed over the case. However that would be difficult (but not impossible to model). The simplest way to model it is to show the weight acting at the centre of mass which is roughly in the middle of the case if the mass is evenly distributed.

20. Oct 19, 2016

I got this. Thank you for help :)