# Tension on a massless string holding a semisphere in place

I am having trouble solving this. The problem is the following:

Suppose a semisphere of radius $$a$$ (with its center of mass at $$h = \frac{3}{8}a$$) of mass $$m$$ located on a sliding surface. A massless, non-extensible string of length $$b$$ (b<a\$) is attached to the edge of the semisphere. What's the tension of the rope in the picture? I am not sure about how to proceed. Should the normal force be pictured straight upwards, or at an angle $$\theta$$ with respect to the vertical axis? My intuition tells me that, if the semisphere is on a sliding surface, the whole body will rotate in the sense that gets the center of mass lower (to the right, in this case) until the torque becomes at equilibrium with the tension of the rope, but I am not sure how to get there analytically.

• How would sliding allow the center mass to be lower? – BowlOfRed Oct 25 '20 at 23:08

## 1 Answer

You can calculate the torque about the point of contact with the floor due to gravity, as well as the torque due to the string. They must counterbalance.