A ramp inclined at 60 degrees with respect to the horizontal rests on a floor. T
ID: 2179275 • Letter: A
Question
A ramp inclined at 60 degrees with respect to the horizontal rests on a floor. The mass of the ramp is 10 kg, and the coefficient of static friction is 0.80. A 5.0 kg block is then placed on the inclined surface of the ramp as shown in the figure. Assume the the friction between the block and the ramp is negligible. What is the magnitude of the frictional force on the ramp? 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)
N1 = m1 g cos = 5 * 9.8 * cos60 = 24.5 N
fs = N1 sin60 = 24.5 * 0.866025 = 21.2176125 21.2 N
b)
fs = N1 sin60 = N2
N2 = m2 g + N1 cos
N1 sin60 = N2
==> m1 g cos60 sin60 = (m2 g + N1 cos)
==> m1 * 9.8 * 0.5 * 0.866025 = 0.8 * (10 * 9.8 + m1 * 9.8 * 0.5)
==> m1 = 0.1885226 * (98 + m1 * 9.8 * 0.5)
==> m1 = 18.4752148 + 0.92376074 m1
==> m1 = 242 Kg
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