Question

A car (mass 1000 kg) is pulling a four-wheeled trailer (mass 200 kg) along a straight flat road using a tow rope. The car is travelling at a constant velocity of 20 m/s.
The coefficient of friction between the road and the tyres of both the trailer and the car is 0.4. There is a force of 21 220 N to the right of the car .
1) Calculate the frictional force acting on the trailer
2) Determine the magnitude of the tension in the tow rope.
3) Calculate the driving force of the car as it pulls the trailer at constant velocity.

Answer 1

1) Since the car-trailer combo is accelerating at 0.4 m/s^2, the net force is

F = m⋅a =(1200kg+400kg)⋅0.4ms^2 = 640 N

The remainder of the 3200N is dealing with the resistances to motion. So friction, air drag, etc. of both vehicles add up to

3200 N − 640 N = 2560 N

That was ignoring the direction of this force. These resistances are forces pointing to the rear. Therefore we need to give this 2560 N a negative sign.

Because the horizontal resistances of the car and the trailer are proportional to their respective masses, the trailer's resistance, ${R}_{t}$, is

${R}_{t}$ = −2560 N × 400 kg / 1200 kg + 400 kg = − 640 N

2) The tension, T in the tow rope is providing force to deal with ${R}_{t}$ and also to provide the acceleration. Note that this is a force pointing forward -- it is encouraging the trailer to keep up with the car.

T = 640 N + ma = 640 N + 400 kg0.4 m/s^2 = 800 N

3) After the tow rope breaks, ${R}_{t}$ will eventually stop it (assuming it stays on the road). ${R}_{t}$ is a force of 640 N and that will provide acceleration ${a}_{t}$ of (solving Newton's 2nd Law for ${a}_{t}$

${a}_{t}$ = ${R}_{t}$ / m = -640 N/400 kg = −1.6 ms^2

${v}^{2}$ = ${u}^{2}$ + 2 × ${a}_{t}$ × d

0² = 25² + 2 × (−1.6) × d

-2 × (-1.6) × d = 25²

d = 625/ 3.2= 2000 m

I hope this helps.