# 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

##### 1 Answer
Jun 14, 2018

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

${F}_{\text{net}} = m \cdot a = \left(1200 k g + 400 k g\right) \cdot 0.4 \frac{m}{s} ^ 2 = 640 N$

The remainder of the 3200N is dealing with the resistances to motion. So friction, air drag, etc. of both vehicles add up to $3200 N - 640 N = 2560 N$. 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

${R}_{t} = - 2560 N \cdot \frac{400 \cancel{k g}}{1200 \cancel{k g} + 400 \cancel{k g}} = - 640 N$

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 ${a}_{t}$

${a}_{t} = {R}_{t} / m = \frac{- 640 N}{400 k g} = - 1.6 \frac{m}{s} ^ 2$

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

${0}^{2} = {\left(25 m {s}^{-} 1\right)}^{2} + 2 \cdot \left(- 1.6 \frac{m}{s} ^ 2\right) \cdot d$

$- 2 \cdot \left(- 1.6 \frac{m}{s} ^ 2\right) \cdot d = {\left(25 m {s}^{-} 1\right)}^{2}$

$d = \frac{625 \frac{m \cancel{^ 2}}{\cancel{{s}^{2}}}}{3.2 \frac{\cancel{m}}{\cancel{{s}^{2}}}} = 2000 m$

I hope this helps,
Steve