This question is from JEE Screening 2005 (Physics)

A block of mass m is at rest under the action of force F against a wall. Which of the following is incorrect? Please see attached file for figure.

a) f = mg [f is frictional force]

b) F = N [where N is normal force]

c) F will not produce torque

d) N will not produce torque

The answer for this question is D. However, I don't fully understand how we can say that F will not produce torque. Under the condition of static equilibrium, all we know is that sum of torques around any line is 0. So, if we take the axis of rotation along centre of mass, then of course the torque by F and mg will be 0. But if we take axis of rotation to be lets say the uppe left corner of the block, then torque by F will not be zero.

When this question appeared in the screening exam, was it clearly mentioned to take the axis of rotation around the centre of mass?

Thanks,

Abhay.

since the dir. of the frictional force=mg is upward for the block to be in eqlm.

also in stable eqlm the net force=0

therefore F=N

since in the fig the F is acting in the line of C.M. so torque by it =0

but N will produce a torque to balance the torque by the frictional force

so N will also shift its position such that it produce anticlockwise torque to balance the clockwise torque by the frictional force

hope u got it.

Animal, like I said before, I already know what you've said, but this solution only make sense when you take the axis of rotation to be through the centre of mass.

It is easy to see that that F=N and f=mg

The condition of static equilibrium is that the sum of torques around any line is 0 (not just centre of mass). So in order for this solution to make sense, the axis of rotation should have been clearly specified...makes sense?

For example, take the axis of rotation to be going through the upper left corner of the cubical block moving in and out of the plane. In this case, the torque by f will be 0 and the other forces will have torques that sum to 0. Torque by F will not be 0 here.

Can anyone understand what I'm saying here?? Was the axis of rotation specified when this problem appeared in JEE.

Thanks,

Abhay.

Wanted to request if anyone could answer my query above?

Thanks.

In this case I think that friction is not coming fully on the side of wall.

mg/2 will come there and remaining mg/2 will come at the surface of application of force F ,isn't it?

Then only moment due to weight(mga) will be balanced by moment due to friction  at F(mg/2*2a) when we take moments wrt the wall.

Even now I am not sure about the answer 'D'. It appears to be wrong!

Once again, just wanted to request if anyone could solve my query above. I think it is essential that someone should be able to explain the answer to a question that appeared on an actual JEE exam. Please take into consideration what I mentioned above when posting the reply. This is not some random question but a question that is very relevant to JEE practice. Any help is really appreciated.

Can any of the forum experts answer this? Anyone?

Thanks.

Hey Folks,

Just wanted to push this question again (please see al the posts above).

Any help would be great.

Thanks.

Hey Varun,

Torque produced by a force depends on the line of rotation. So as I said previously, if you take the axis of rotation to be through the centre of mass, only then the given solution makes sense.

The answer given above is same as the one posted on the solutions given by Fiitjee, Narayana or Arihant previous year's solutions.

My question is, if we are just supposed to assume what axis of rotation to take. Any hints or suggestions on this?

I'm surprised none of the forum experts have been able to comment on this. Common guys, have a dig at this.