It\'s an icy day, and a car (m=1000kg) has skidded off the road down a 25 degree

Question

It's an icy day, and a car (m=1000kg) has skidded off the road down a 25 degree embankment. To make matters worse, the cars brakes have locked up, so the wheels can't turn. A tow truck (m=8000kg) has come along to pull the car out, but the road is still icy. There is a pulley system (assume frictionless) available that allows the truck to pull sideways, while the chain (of negligible) attached tothe car pulls parallel up the embankment. If the coefficient of kinetic friction on the embankment is Mk=0.25, what is the minimum coefficient of static friction between the icy road and the two truck that will allow the tow truck to pull the car out of the embankment? Don't worry about the static friction that would need to be overcome to get the car moving, and start by drawing free body diagrams for the truck and the car.

Explanation / Answer

This is a wordy question. Assuming I read it correctly, First, the car is on an angle 25 degrees. A rope will pull up, but gravity and friction hold it down. At a minimum, we need to overcome gravity and friction. So parallel to the hill, Force(rope) = Force(gravity on hill) + Force(friction of hill). Force(rope) = mgsin25 + (Mk)(Normal force). The normal force is equal and opposite the y component of gravity on the car while on the hill, so Normal Force = mgcos25 = (1000)(9.8)cos25 = 8881.8 N. Plug this into the above equation to find the Force(rope) = 6362.11 N. Now, the tow truck needs to pull with a MINIMUM tire force of 6362.11 to overcome the rope holding it down, so Force(friction of tires) = (Mstatic truck)*(Normal force truck), so 6362.11 = Mstatic * mg. Mass of truck is 8000 kg, so Mstatic = .0811. This answer seems low, let me know if its correct.

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