A car has a width of x=1.60m and a height y=1.20m; you may assume the center-of-

Question

A car has a width of x=1.60m and a height y=1.20m; you may assume the center-of-mass is in the middle of the car. You lift the right side using a simple jack, and for simplicity let’s assume the car is a rectangle with its axis of rotation at the bottom left as shown. The jack is set up so it makes an angle of theta=35° above the horizontal and provides a normal force of F=15000 N to the bottom-right of the car. The center of mass, the jack, and the pivot point are in the same plane, shown as a rectangle in the figure.This is a two-dimensional problem; ignore the length of the car.

a) If the applied force is just enough to start to lift the car, what is its mass?

b) What is the minimum coefficient of friction the tire must have with the road to prevent it from sliding?

Explanation / Answer

This is a question of torque balance

a) Taking moment about the pivot point

mg * x/2   = Fsin 35 * x

where m = mass of the car

mg/2 = Fsin 35

m = 2Fsin 35 / g

m = 2*15000*sin 35 / 9.8

m = 1755.84 kg

so the mass of the car is 1755.84 kg

b) friction force will act in +x direction

friction force = Fcos 35

u *N = Fcos 35

u (mg) =F cos 35

u *(1755.84*9.8) = 15000 cos 35

u = 0.714

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.