A uniform sphere of radius R=0.2m and M=3kg held at rest on incline plane with angle=30deg. The top of the sphere is connect by string which also connected to incline.
what is tension on string, what is normal force which act from incline to sphere and what is frictional force?

Question

A uniform sphere of radius R=0.2m and M=3kg held at rest on incline plane with angle=30deg. The top of the sphere is connect by string which also connected to incline.
what is tension on string, what is normal force which act from incline to sphere and what is frictional force?

anonymous · Answer

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ivancsc1996 · Answer

Well the  string in that picture is clearly doing a torque that is being contarested by the torque that the friction is doing. So the sumatory of torques would be$$\tau _{string}+ \tau _{friction}=0 \rightarrow \tau _{string}= \tau _{friction}$$(where tau is torque)$$\tau _{string }=TR$$(where T is tention and R is radius)$$\tau _{friction}=uNR$$(where u is the coefficient of friction, N is the normal force and R is the radius)

Well we know the normal force must be the force excerted due to gravity by the sphere to the inclined plane.$$N=Mgcos(\theta)$$(where M is mass, g is the gravitational acceleration and theta is the angle of inclination) The problem is that I we know have three variables and only two equations. Maybe we need an equation involving moment of inertia. $$I= \frac{ 2 }{ 5 }MR ^{2}$$

anonymous · Answer

Well, I cant see that moment of inertia can give us something. I agree with all outer staff but it is not give us idea about u and T.
Really i have now idea where to move to solve this.

vincent-lyon.fr · Answer

The result does not depend on R.

$\tau _{friction}=\mu NR$ is not correct: as the system does not slip, there is no equality between force of friction and $\mu N$

Your missing equation is that for rotational equilibrium. It is very simple to write down: the result is simply that tension in the string equals friction force.
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