Question

A 22 kg toboggan is pulled along at a constant speed by a horizontal force of 31N. a) What is the force of gravity on the toboggan? b) What is the coefficient of friction? c) How much force is needed to pull the toboggan if two children each with a mass of 50 kg are sitting on it?

Answer 1

(a) This is essentially asking for the weight of the toboggan, which is

w = (22 kg) (9.80 m/s²) = 215.6 N ≈ 220 N

(b) Being pulled with constant speed means the toboggan is in equilibrium, so

n + (-w) = 0

where n is the magnitude of the normal force, and

31 N + (-f ) = 0

where f is the magnitude of the friction force.

f is proportional to n by a factor of µ such that

f = µ n

We have

n = w = 215.6 N

f = 31 N

so

31 N = µ (215.6 N) → µ ≈ 0.14

(c) Let p denote the magnitude of the required pulling force. The combined mass of the toboggan and children is now 122 kg, so its weight is

w = (122 kg) (9.80 m/s²) = 1195.6 N

which means the normal force is also n = 1195.6 N.

We want to have

p + (-f ) = 0 → p = f

so we just need to find the magnitude of the friction force using the coefficient from part (b).

p = f = 0.14 (1195.6 N) ≈ 170 N