Find the range of P for which the P for which the weight rem

Find the range of P for which the P for which the weight remains at rest. Take mu_3=0.3 for all the wheels each of which has radius of 1.5 inches

Solution

Let the weight have a tendency to go down.

then T - P = mu*R
let theta be the angle between the right string of the leftmost pulley and the normal passing throught the pulley
theta = arctan(4/9) = 23.96 deg
so half angle = (90+23.96)/2 = 56.98 deg
R = [P+T]cos(56.98)
T-P = mu[P+T]cos(56.98)
T = P(1 + mucos(56.98))/(1 - mu(56.98)) = 1.39P

similiarly half angle for next pulley = [23.96 + arctan(7/8)]/2 = 32.57 deg
so T\' = T(1 + mucos(32.57))/(1 - mucos(32.57)) = 1.39P*1.6767 = 2.18P

similiarly, half angle for last pulley = 41.18/2 = 20.9 deg

Now W = T\'cos(20.9) + T\"
and R + Wcos(20.9) = (T\"+T\')cos(20.9)
and T\" -T\' = mu R
or T\" = (mu*cos(20.9)(T\' -W)+ T\')/(1 - mucos(20.9)) = (0.28*(2.18P - W) + 2.18P)/0.72 = 0.389(2.18P - W) + 3.0277P = 3.875P - 0.389W

but T\" = W - T\'*cos(20.9) = W - 2.18Pcos(20.9) = W - 2.03P = 3.875P - 0.389W
1.389W = 5.91P
W = 4.28P -- (1)

Let the weight have a tendency to go up.

then P - T = mu*R
let theta be the angle between the right string of the leftmost pulley and the normal passing throught the pulley
theta = arctan(4/9) = 23.96 deg
so half angle = (90+23.96)/2 = 56.98 deg
R = [P+T]cos(56.98)
P - T = mu[P+T]cos(56.98)
T = P(1 - mucos(56.98))/(1 + mu(56.98)) = 0.719P

similiarly half angle for next pulley = [23.96 + arctan(7/8)]/2 = 32.57 deg
so T\' = T(1 - mucos(32.57))/(1 + mucos(32.57)) = 0.71P*0.59 = 0.423P

similiarly, half angle for last pulley = 41.18/2 = 20.9 deg

Now W = T\'cos(20.9) + T\"
and R + Wcos(20.9) = (T\"+T\')cos(20.9)
and T\' -T\" = mu R
or T\" = (-mu*cos(20.9)(T\' -W)+ T\')/(1 + mucos(20.9)) = (-0.28*(0.423P - W) + 0.423P)/1.28 = -0.21875(0.423P - W) + 0.33P = 0.237P + 0.21875W

but T\" = W - T\'*cos(20.9) = W - 0.423Pcos(20.9) = W - 0.395P = 0.237P + 0.21875W
0.21875W = 0.165P
W = 0.755P -- (2)

Range of weight -> [0.755P,4.28P]