Calculate the torque exerted by the car around the back wheels

[stoopid]

1.A mechanic jacks up a car to an angle of 7.0° to change the front tires. The car is 2.90 m long and has a mass of 1180 kg. Its center of mass is located 1.12 m from the front end. The rear wheels are 0.40 m from the back end. Calculate the torque exerted by the car around the back wheels.


2.Torque=Fd(sin feta)
F=mg



3.d=2.9m-1.12m-.4m
= 1.38m

F=1180kg(9.8m/s^2)=11564N

T=11564N(1.38m)sin (7)

Is this correct?

 

[dynamicsolo]

1. A mechanic jacks up a car to an angle of 7.0° to change the front tires. The car is 2.90 m long and has a mass of 1180 kg. Its center of mass is located 1.12 m from the front end. The rear wheels are 0.40 m from the back end. Calculate the torque exerted by the car around the back wheels.

I am presuming that they mean the torque exerted by the weight of the car.

d=2.9m-1.12m-.4m
= 1.38m

F=1180kg(9.8m/s^2)=11564N

T=11564N(1.38m)sin (7)

I will make two comments. The weight of the car is probably supposed to be treated as if it acts from the center of the body of the car. How far away is the midpoint of the car from the back wheels? Also, the angle in the torque equation is measured between the direction of the moment arm and the direction of the relevant force. The moment arm would point from the back wheels to the midpoint of the car (I'd suppose). Which way does the weight force for the car point? What angle does that make to the moment arm?

 

[dynamicsolo]

The weight would point down and it would make a angle of 83 degrees.

The angle I'm describing is between the vector going from the rear wheel to the center of mass, which points left and 7º above the horizontal, and the weight vector, which points straight down. That angle would be 97º. However, sin 97º = sin 83º, so either gives the same result.

I think the answer to your first comment is 1.38m.

Excuse me; yes, that's right. I overlooked that information when I start typing. Thanks for the diagram -- that's definitely helps.