Question

The rope connecting the 6-kg block A with block B passes over a fixed cylinder. Determine the largest and smallest masses of block B for which static equilibrium is possible if the coefficient of static friction is 0.30.

Answer 1

when system is in equilibrium, ratio of tension on two sides of string is same as ratio of weights on two sides.
hence T1/T2 = Ma/Mb

T2-Mag

Consider small element of string over the pulley as shown in the figure. Tension difference across this element
dT = $\mu N$   where $\mu$ is static friction coefficient and N is normal force between string and pulley at thos position.
If T is tension at this position
N = 2T (d $\theta$ /2 ) ( resultant of tension from two ends is balanced by normal force)
From these two equations we get
dT = $\mu$ T d$\theta$
dT /T = $\mu$ d$\theta$
Integrating from T1 to T2 on left side and 0 to pi on right side we get
Ln(T2/T1) = $\mu$ pi
T2/T1 = Ma/Mb, if A is heavier
Ln( 6 /Mb) = 0.3 pi
Mb = 2.34 kg
T2/T1 = Mb/Ma , if B is heavier
Ln( Mb/6) = 0.3 pi
Mb = 15.39 kg
so smallest and largest possible avlues of B are 2.34 and 15.39 kgs, for static equilibrium