Question

A small 5.7 -kg box is stationary on top of a ramp. The ramp makes an angle β with the horizontal. The coefficient of kinetic friction is 0.17 and the coefficient of static friction is 0.36. Find the minimum angle β (in degrees) where the box starts to move.

Answer 1

From Free body diagram,

By force balance in perpendicular direction of ramp,

N = m*(g*cos)

By force balance in direction of ramp,

fr = m*(g*sin)

where, fr = friction force = *N

for minimum angle, = coefficient of static friction = s = 0.36

given, m = 5.7 kg

So, s*m*g*cos = m*g*sin

tan = 0.36

= arctan(0.36)

= 19.8 deg

So, box starts to move.when angle of ramp is 19.8 deg.

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