A box is resting on a rough inclined surface. The coefficient of kinetic frictio

Question

A box is resting on a rough inclined surface. The coefficient of kinetic friction has a value of, ?s =0.75. The coefficient of kinetic friction has a value of, ?k=0.5. The rough inclined surface is moveable (as the diagram indicates) and the incline is able to be moved through the angles of ?=0 to ?=60 degrees. The mass of the box is 1.0kg and you can use g=10 for the value of gravity. a) draw a Free Body Diagram of the box as it rests on the incline at some arbitrary angle ?. b)......the incline is moved from zero degrees up until an angle at which the box just begins to slip, lets call this angle the "slip angle", ?slip. i) Determine the value of ?slip ii) After the slippage occurs the box accelerates down the plane, determine the value of this acceleration.

Explanation / Answer

b) a = g sin ? - µs g cos ? = 0 g sin ? = µs g cos ? sin ? = µ s cos ? sin ? / cos ? = 0.75 = tg ? ? = 36.87 ° c) For an incline at angle 36.387°degrees and coefficient of friction µk=0.5, the acceleration of the object down the incline is: a = g sin ? - g µk cos ? = g ( sin? - µk cos? ) = 10 ( 0.6 - 0.5 * 0.8 ) = 10 * (0.6 - 0.4) = 2 Acceleration =a = 2 m/s2

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