12.Two masses are arranged as shown on a ramp that makes an angle of 36° with the horizontal. Mass m1 is twice mass m2 . What is the minimum coefficient of static friction required between mass m1 and the ramp to ensure equilibrium?

"1.7456" would be the minimum coefficient of static friction required.

Given:

Angle,

Mass,

To keep the system in equilibrium forces acting on mass "" must balance forces acting on "".

That's,

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or,

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Thus,

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hence,

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Thus the above answer is right.

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Force  M2 exerts on tension T

T = M2 g

T = M1 g sin theta - M1 g u cos theta

Note that M1 g is exerted down the plane while the frictional force opposes this motion and will be directed in the opposite direction of M1 g

M2 g = M1 g sin theta - u M1  g cos theta

1 = 2 sin 36 - u 2 cos 36

.5 = sin 36 - u cos 36

u = (sin 36 - .5) / cos 36 = .108

Any friction less that this will result in sliding

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