Posted by Bella on Thursday, November 4, 2010 at 8:32pm.
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.
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.
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