A block of mass m, acted on by a force of magnitude F directed horizontally to t

Question

A block of mass m, acted on by a force of magnitude F directed horizontally to the right as shown above, slides up an inclined plane that makes an angle theta with the horizontal. The coefficient of sliding frction between the block and the plane is mu.
Develop an expression in terms of m, theta, F, mu, and g for the block's acceleration up the plane.
Develop an expression for the magnitude of the force F that will allow the block to slide up the plane with constant velocity.
What relation must theta and mu satisfy in order for this solution to be physically meaningful?

Explanation / Answer

N = Fsin + mgcos

friction force f = N = (Fsin + mgcos)

Fcos - mgsin - (Fsin + mgcos) = ma

a = F(cos - sin) - mg(sin + cos) / m ..........Ans

block will slide with constant velocity when a = 0

so , F(cos - sin) - mg(sin + cos) = 0

   F =mg(sin + cos) / (cos - sin)   .............Ans

for this solution to be meaningful ,

cos should not be equal tp sin

so must not be equal to tan-1(1/) .

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