Question

A car of mass 1200kg pulls a trailer of mass of 400kg along a straight horizontal road. The car and trailer are connected by a tow-rope modelled as a light inextensible rod. The engine of the car provides a constant driving force of 3200N. (see ? below)

The horizontal resistances of the car and the trailer are proportional to their respective masses. Given that the acceleration of the car and trailer is 0.4ms-2,
a) find the resistance to motion on the trailer
b) find the tension in the tow-rope
When the car and trailer are travelling at 25ms-1 the tow-rope breaks. Assuming the resistances to motion are unchanged,
c) find the distance the trailer travels before coming to a stop

Answer 1

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

`F_"net" = m*a = (1200 kg+400 kg)*0.4 m/s^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 `3200N - 640N = 2560N`. 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 = -2560N*(400 cancel(kg))/(1200 cancel(kg)+400 cancel(kg)) = -640 N`

b. 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

c. 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 m/s^2`

`v^2 = u^2 + 2*a_t*d`

`0^2 = (25ms^-1)^2 + 2*(-1.6 m/s^2)*d`

`-2*(-1.6 m/s^2)*d = (25ms^-1)^2`

`d = (625 (mcancel(^2))/cancel(s^2))/(3.2 cancel(m)/cancel(s^2)) = 2000 m`

I hope this helps,
Steve