- #1

Matt Raj

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- Homework Statement
- A horizontal stick of mass m has its left end attached to a pivot on a plane inclined at an angle θ, while its right end rests on the top of a cylinder also of mass m which in turn rests on the plane, as shown in Fig. 2.36. The coefficient of friction between the cylinder and both the stick and the plane is μ.

(a) Assuming that the system is at rest, what is the normal force from the plane on the cylinder?

- Relevant Equations
- T=rxF

Ff=µ*N

The friction from the rod and the friction from the plane on the cylinder should be the same due to torque equilibrium on the cylinder. If we let N_1 be the normal force on the rod and N_2 be the normal force on the cylinder from the plane, I expected µN_1 = µN_2. Looking at torque on the rod, mg/2=N_2. Thus N_1 = mg/2, but this clearly isn't correct. The normal force on from the plane should be greater than N_1 and the correct answer is 3mg/2 which I got from writing the equations only in terms of equal frictions and not using the equation µN_1 = µN_2 and instead using F_1=F_2, where F_1 is the friction from the rod and F_2 is the friction from the plane.

Is the reason my first method was wrong because the given µ in the problem is simply the maximum possible value for friction and in this problem, the frictions are equal but both have different normal forces and µ values less than the given µ?

Is the reason my first method was wrong because the given µ in the problem is simply the maximum possible value for friction and in this problem, the frictions are equal but both have different normal forces and µ values less than the given µ?