Question

1. A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 300 N and a vertical force P are then applied to the block. The coefficient of friction for the block and surface are s-0.400 and 4-0.200. a) Determine the magnitude of the frictional force acting on the block if the magnitude of Pis 60.0 N. (8 points) b) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 80.0 N. (8 points)

Answer 1

1

a)

N = Normal force acting in upward direction

P = vertical force acting in upward direction = 60 N

m = mass of the block = 20 kg

W = weight of the block in downward direction = mg = 20 x 10 = 200 N

Using equilibrium of force in vertical direction , we get

P + N = W

60 + N = 200

N = 140 N

Maximum static frictional force possible for the block is given as

f_smax = $\mu _{s}$ N = (0.4) (140) = 56 N

F = applied force in horizontal direction = 50 N

Since the maximum static frictional force is greater than the applied force, hence static frictional force value adjust itself to the value of applied force, hence

f_s = static frictional force on the block = F = 50 N

f_s = 50 N

b)

N = Normal force acting in upward direction

P = vertical force acting in upward direction = 80 N

m = mass of the block = 20 kg

W = weight of the block in downward direction = mg = 20 x 10 = 200 N

Using equilibrium of force in vertical direction , we get

P + N = W

80 + N = 200

N = 120 N

Maximum static frictional force possible for the block is given as

f_smax = $\mu _{s}$ N = (0.4) (120) = 48 N

F = applied force in horizontal direction = 50 N

Since the maximum static frictional force is smaller than the applied force, hence block moves and kinetic frictional force acts on it which is given as

f_k = kinetic frictional force on the block = $\mu _{k}$ N = (0.2) (120) = 24 N

f_k = 24 N