Question

1. repo. a fancy car (mass = 1500 kg) is connected via a chain to a tow truck and is towed up a hill at a constant velocity. the slope of the hill is 20°. friction is negligible.

a) draw a free - body diagram for the car, and calculate the tension in the chain.
b) approximately, what is the mechanical advantage created by the incline? (compare the force to if you were to pull the car straight up.)

Explanation:

Step1: Analyze forces on car

The forces acting on the car are its weight $mg$ (where $m = 1500$ kg and $g=9.8$ m/s²), the normal force $N$ perpendicular to the slope and the tension $T$ in the chain along the slope. Since the car moves at a constant velocity, the net - force along the slope is zero.

Step2: Resolve weight along slope

The component of the weight along the slope is $mg\sin\theta$. In the direction along the slope, $T - mg\sin\theta=0$ (because $F_{net,x}=0$ for constant velocity motion), so $T = mg\sin\theta$.
Substitute $m = 1500$ kg, $g = 9.8$ m/s² and $\theta = 20^{\circ}$ into the formula:
$T=1500\times9.8\times\sin(20^{\circ})$
$T = 1500\times9.8\times0.342$
$T=4998.6\approx5000$ N

Step3: Calculate mechanical advantage

The force required to lift the car straight up is $F_{lift}=mg=1500\times9.8 = 14700$ N. The force required to pull the car up the incline is $T = mg\sin\theta$.
The mechanical advantage $MA=\frac{F_{lift}}{T}=\frac{mg}{mg\sin\theta}=\frac{1}{\sin\theta}$
Substitute $\theta = 20^{\circ}$, $MA=\frac{1}{\sin(20^{\circ})}=\frac{1}{0.342}\approx2.92$

Answer:

a) The free - body diagram shows the weight $mg$ acting downwards, the normal force $N$ perpendicular to the slope and the tension $T$ along the slope. The tension in the chain is approximately $5000$ N.
b) The mechanical advantage created by the incline is approximately $2.92$.