Determine the magnitude and direction of the friction force

Determine the magnitude and direction of the friction force which the vertical wall exerts on the 45 kg block if theta = 15 degree and theta = 30 degree.

Solution

Reaction from the wall = 500*cos theta

Component of force along the wall = 500*sin theta

Weight of the component = mg = 45*9.81 = 441.45 N

a)

When theta = 15 deg,

Reaction from the wall = 500*cos 15 = 482.96 N

Static friction force = 0.5*482.96 = 241.48 N

Kinetic friction force = 0.4*482.96 = 193.18 N

Component of force along the wall = 500*sin 15 = 129.41 N

Net downward force = 441.45 - 129.41 = 312.04 N

Since 312.04 > 241.48, the block will tend to move downwards. Hence, friction force of 193.18 N will act upwards.

b)

When theta = 30 deg,

Reaction from the wall = 500*cos 30 = 433.01 N

Static friction force = 0.5*433.01 = 216.51 N

Kinetic friction force = 0.4*433.01 = 173.21 N

Component of force along the wall = 500*sin 30 = 250 N

Net downward force = 441.45 - 250 = 191.45 N

Since 191.45 < 216.51, the block will not move. Hence, friction force = 191.45 N acting upwards.