Question
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.
Answers
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)
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)
this is best question
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