Question

o m all, 2019 Problem 1 The USA Bobsledding team is training for the Olympics by pushing a large box across a gymnasium floor. The box has a mass of 135 kg. The coefficient of static friction between the floor and box is 0.6, and the coefficient of dynamic friction between the box and floor is 0.45. (12.5 pts a) (2.5 pts) What is the maximum frictional force acting on the box when it is not moving? b) (2.5 pts) What is the frictional force acting on the box when it is moving? c) (2.5 pts) If the net force applied to the box is directed completely horizontal and the box is not moving, what is the minimum force that must be applied to cause the sled to move? (2.5 pts) If the net force applied to the box is directed completely horizontal and the box is already moving, what is the minimum force that must be applied to keep the sled moving? (2.5 pts) If this were a real training run on ice, the 2 bobsledders would jump in the sled. Assume the sled has a mass of 135 kg. Now the entire mass is the mass of the sled plus the total mass of the bobsledders. If the total mass of the bobsledders was 220 kg, and the coefficient of static friction between the sled and ice was 0.19 and the coefficient of dynamic friction between the sled and ice was 0.04, what is the frictional force between the sled and ice as it is moving?

Answer 1

a) When the object is not moving, the friction is static friction.

The frictional force acting on the box is given by

$F_s = \mu_s * N = mg\mu_s$(N is the normal reaction force defined by newton's third law)

Here, m = 135 kg

g = 9.8 m/s²

$\mu_s$ = 0.6

So,

$F_s = mg\mu_s = 135*9.6*0.6 = 777.6N$

b) When the object is moving, the friction is dynamic friction.

The frictional force acting on the box is given by

$F_d = \mu_d * N = mg\mu_d$

$\mu_d$ = 0.45

So,

$F_d = mg\mu_d = 135*9.6*0.45 = 583.2N$

c) When the object is at rest, to make it move, a force equal to the static frictional force has to be applied horizontally.

So,

d) If the object is already moving, the static friction is already overcome and a minimum force equal to the dynamic friction of the mass is to be applied horizontally to maintain the body in motion.

e) Here, the object is already moving.

$F_d = mg\mu_d$

Here, m = 135+220 = 355 kg

$\mu_d$ = 0.04

$F_d = mg\mu_d = 355*9.8*0.04 = 139.16 N$