Question

Block A is placed on a rough wedge B which is
placed on a smooth surface. The wedge has angle
of inclination of 53° and is imparted a horizontal
acceleration g towards right. Block A is given an
initial velocity v, with respect to wedge. Find the
coefficient of friction for which block A moves
with constant velocity v, with respect to wedge.
(g =10 m/s)​

Answer 1

Answer:

Explanation:

As shown in fig.

ma is pseudo force towars left.

If block A moves with constant velocity then net force is zero.

Force along the surface of bock B is;

macosθ=f+mgsinθ ..................(1)

where f− frictional force

force opposite to the movement of block A

Let μ= coefficient of frictional force

N=masinθ+mgcosθ

(1) becomes;

macosθ=μ(masinθ+mgcosθ)+mgsinθ

where a=g,θ=30

mgcos30 = μ(mgsin30+mgcos30) + mgsin30

g ( √3/2) = μ ( g/2 + √3g/2 ) + g/2

μ [ (1 + √3) / 2] = (√3 - 1)/2

μ = (√3 - 1)/2