A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force vector F of magnitude 6.5 N and a vertical force vector P are then applied to the block. The coefficients of friction for the block and surface are Mus = 0.43 and Muk = 0.22, (a) Determine the magnitude of the frictional force acting on the block if the magnitude of vector P is 8.0 N. N (b) Determine the magnitude of the frictional force acting on the block if the magnitude of vector P is 10.0 N. N (c) Determine the magnitude of the frictional force acting on the block if the magnitude of vector P is 12.0 N.

Explanation / Answer

assuming the box will not move ,

a) friction force , f = mu(s)* N
f = 0.43*( mg-P) = 0.43*(2.5*9.8-8) = 7.095
as this is more than F so ,box will not move and so, the friction acting will be just equalt to F

actual friction force ,f = F
f =6.5 N

b) friction force , f = mu(s)* N
f = 0.43*( mg-P) = 0.43*(2.5*9.8-10) =6.235 N
as this is less than F so ,box will move and so, the friction acting will be just equalt to mu (k) * N
so ,

f = 0.22*( 2.5*9.8-10) = 3.19 N

c)
in this case also ,
box wil move.
so,

f = 0.22*( 2.5*9.8-12) =2.75 N

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