A block of mass M resting on a 21.5° slope is shown. The block has coefficients of friction μs=0.605 and μk=0.344 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.57 kg. What is the minimum mass M1 that will stick and not slip?

Question

bobpursley · Answer

well, tension is 2.57g up the slide
static friction is .605*M1*g*cosTheta, down the slide
gravity down the slide=M1*g*sinTheta

so,

2.57g=.605M1*g*cosTheta-M1*g*sinTheta

solve for M1

check my thinking and typing.

Scotty · Answer

Bob was mostly right, however when he took into consideration of the gravity on the down side he forgot to input the static friction that goes against gravity pulling the block down the slope.

Tension= (Us*(mass*9.8cos(theta))-((mass*9.8sin(20))-(Us*(mass*9.8cos(20))

So without numbers and only variables:
T=(Us*N)-(g)-(Us*N)

obviously, when you have a slope, you have to factor the theta in, as I did in the first equation.