Hemisphere being pulled by a rope problem

[brianws33]

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 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" referring 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.

 

[thomate1]

I would first clarify the problem statement and make sure that all the given information is accurate and complete. This includes confirming the units of the given variables and determining if any assumptions need to be made.

Next, I would approach the problem by using the principles of rotational motion and equilibrium. The hemisphere is being pulled at a constant velocity, so we can assume that the sum of the forces acting on it is equal to zero. This means that the force applied, F, is equal to the force of friction, Ff.

To determine the angle at which the hemisphere is inclined, we can use the fact that the center of mass is at a distance of 3R/8 from the flat surface. This means that the line of action of the applied force must pass through this point. We can then use basic trigonometry to find the angle at which the hemisphere is inclined.

As for the use of torques, they may not be necessary in this problem as we are dealing with a constant velocity and not a rotational acceleration. However, we can still use the concept of torque to confirm our solution. The torque caused by the applied force is equal to the force multiplied by the perpendicular distance from the axis of rotation. In this case, the axis of rotation can be chosen to be at the point of application of the force, and the perpendicular distance can be calculated using the angle we found earlier. If the torque caused by the applied force is equal to the torque caused by the friction force, then our solution is correct.

In conclusion, the key steps in solving this problem would be to confirm the given information, use the principles of rotational motion and equilibrium, and use basic trigonometry to find the angle at which the hemisphere is inclined. The use of torques can also be helpful in confirming the solution.