# Hemisphere being pulled by a rope problem

## Homework Statement

A hemisphere of radius R and mass M, curved side down, is pulled along a plane at a constant velocity by
means of a horizontal force F. The coefficient of sliding friction is mu. Find the the angle at which the hemisphere
is inclined as it is pulled. C.M. of the hemisphere is at a distance 3R/8 from it's flat surface. The force applied is
at the equator of the hemisphere.

## Homework Equations

Sum of the forces = ma =0
Sum of the Torques = Iw =0 (possibly)

## The Attempt at a Solution

I determined a force diagram with the standard normal force equal to weight and Ff equal to the force applied.
Then we determined to use sum of the torques being equal to zero. Problem is is that we have all but the applied force
at the axis of rotation so it is equal to zero and we have the rotational normal force and rotational force of gravity canceling out with the
friction force being equal to zero if you set the axis of rotation at point of applied force.

Are the sum of the torques equal to zero?

Are the torques placed in the right position?

Should we use torques at all?
Any suggestions would help immensely.

nrqed
Homework Helper
Gold Member

## Homework Statement

A hemisphere of radius R and mass M, curved side down, is pulled along a plane at a constant velocity by
means of a horizontal force F. The coefficient of sliding friction is mu. Find the the angle at which the hemisphere
is inclined as it is pulled. C.M. of the hemisphere is at a distance 3R/8 from it's flat surface. The force applied is
at the equator of the hemisphere.

## Homework Equations

Sum of the forces = ma =0
Sum of the Torques = Iw =0 (possibly)

## The Attempt at a Solution

I determined a force diagram with the standard normal force equal to weight and Ff equal to the force applied.
Then we determined to use sum of the torques being equal to zero. Problem is is that we have all but the applied force
at the axis of rotation so it is equal to zero
what is "it" refering to ?

There is no specific axis of rotation since the object does not rotate. You can pick any axis you want for the axis of rotation and the net torque with respect to that axis must be zero.
and we have the rotational normal force and rotational force of gravity canceling out
there is no need to use the word "rotational". Yes, the normal force cancels out the weight.
with the
friction force being equal to zero if you set the axis of rotation at point of applied force.

I don't understand this part. The friction force is not zero. And if you set the axis of rotation at the point of the applied force , the torque produced bythe friction force is not zero either.

Yes, the sum of the torques is zero and the sum of the forces is also zero. Pick any point you want for the axis of rotation and impose that the net torque with respect to that axis must be zero. Using in addition that the net force along X is zero and that the net force along Y is zero should give enough information.