A box of mass m=10 kg rests on a surface inclined at 0=30° to the horizontal. It is connected by a light weight cord which passes over a massless and frictionless pulley to a second box of mass Mb which hangs freely . If the coefficient of static friction is 0.40 determine what range of values for mass Mb will keep the system at rest.

Question

Anonymous · Accepted Answer

first when will it slide down?
component of weight down slope = m g sin 30
normal force = m g cos 30
friction force up = 0.4 m g cos 30
other weight force up = Mb g
so for force up = force down
Mb g + 0.4 m g cos 30 = m g sin30
or Mb = 10 (sin 30 - .4 cos 30)
now
for when it slides up reverse friction effect
Mb g - 0.4 mg cos 30 = m g sin 30
or Mb = 10 (sin 30 + .4 cos 30)

werksew · Answer

this is best question