A ramp inclined at 60.0 degrees with respect to the horizontal rests on a floor.

Question

A ramp inclined at 60.0 degrees with respect to the horizontal rests on a floor. The mass of the ramp is 10.kg, and the coefficent of static friction between the bottom of the ramp and the floor is .8. A 5.0 kg block is then placed on the inclined surface of the ramp. Assume that the ramp remains stationary with respect to the floor and that friction between the ramp is negligible. What is the magnitude of the frictional force? Suppose a block with a different mass were placed on the ramp. If the mass of this block were large enough, the frictional force would not be enough to keep the ramp stationary. What is the maximum the mass of the block could be if the ramp is to remain stationary?

Explanation / Answer

(a) magnitude of the Frictional Force = mgcos60 = 5 * 10 * cos30 = 43.3 N. (b) The Frictional Force Between the block and ramp = mgcos30 = m * 8.66 ; The vertical component of the Frictional Force + normal reaction on ramp by the block = m * 10 * cos30 Which is equal to Static friction between ramp and floor ===> m * 10 * cos30 = 0.8 * (10 + m) * g ===> 8.66 * m = 8 * (10 + m) ===> m = 121.17 kg

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