A mechanic jacks up a car to an angle of 10◦ with the horizontal in order to change the front tires. The car is 2.78 m long and has a mass of 1110 kg. Its center of mass is located 1.15 m from the front end. The rear wheels are 0.25 m from the back end.
around the back wheels. The acceleration of gravity is 9.81 m/s2 .
Answer in units of N · m
Calculate the torque exerted by the car around the back wheels. The acceleration of gravity is 9.81 m/s2 .
Answer in units of N · m
To calculate the torque exerted by the car around the back wheels, we need to find the force and the lever arm.
1. Finding the force (F):
The force can be calculated using the formula F = mg, where m is the mass of the car and g is the acceleration due to gravity.
Given: mass (m) = 1110 kg, acceleration due to gravity (g) = 9.81 m/s^2
F = 1110 kg × 9.81 m/s^2 = 10899.9 N (approximately 10899.9 N)
2. Finding the lever arm (r):
The lever arm is the perpendicular distance between the axis of rotation (the back wheels) and the line of action of the force. In this case, the lever arm is the distance between the center of mass of the car and the back wheels.
Given: length of the car (L) = 2.78 m, center of mass distance from front end (d) = 1.15 m, distance of rear wheels from back end (w) = 0.25 m
Lever arm (r) = L - (d + w) = 2.78 m - (1.15 m + 0.25 m) = 1.38 m
3. Calculating the torque (τ):
The torque is calculated using the formula τ = F × r.
Given: F = 10899.9 N, r = 1.38 m
τ = 10899.9 N × 1.38 m = 15055.862 N · m (approximately 15055.862 N · m)
Therefore, the torque exerted by the car around the back wheels is approximately 15055.862 N · m.