How Do You Calculate Torque in Mechanical Situations?

[hshphyss]

1.) A mechanic jacks up a car to an angle of 8.0° to change the front tires. The car is 3.20 m long and has a mass of 1160 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'm not sure how to set this up. I know the equation for torque is t=fr sin(theta)
2.) A bucket filled with water has a mass of 71 kg and is attached to a rope that is wound around a 0.035 m radius cylinder. A crank with a turning radius of 0.15 m is attached to the end of the cylinder. What minimum force directed perpendicularly to the crank handle is required to raise the bucket?
--I think this is a moment of inertia problem... so for that shape it would be MR^2 but after you find that what would you plug that into?
Thank-you

 

[lightgrav]

1) "r" points from the axis of rotation (the back wheels) to the point where the Force is applied. The first Force that you need to consider is the Force applied TO the car BY THE EARTH (gravity), which is downward. The angle that you take the sin of (in "sin theta") is the angle from the "r" direction to the "F" direction (swept thru by your fingers as they more from "along r" to "along F". In your case, it is MORE than 90 degrees.

[the wording "torque applied BY the car around the back wheels" implies that
there is some OTHER object, besides the car, that this torque is applied TO. I can't think of any ... the car cannot apply a torque to itself!]

2) the MINIMUM Force needed to raise the bucket will have zero acceleration of the bucket, and zero angular acceleration of the drum.
So you do NOT need to compute the moment of rotational Inertia.

 

[quicknote]

for reaching out for help with torque problems. I can provide you with some guidance on how to approach and solve these types of problems.

Firstly, it is important to understand the concept of torque. Torque is a measure of the rotational force applied to an object and is calculated by multiplying the force applied by the distance from the pivot point. In your first problem, the pivot point is the back wheels of the car.

To solve this problem, you can use the equation you mentioned: t=fr sin(theta). In this equation, t represents torque, f represents the force applied, r represents the distance from the pivot point, and theta represents the angle at which the force is applied. To find the torque exerted by the car around the back wheels, you will need to calculate the force applied and the distance from the pivot point.

To find the force applied, you can use the equation F=ma, where m is the mass of the car and a is the acceleration due to gravity (9.8 m/s^2). The distance from the pivot point can be calculated by subtracting the distance from the center of mass to the front end from the total length of the car.

Once you have these values, you can plug them into the torque equation and solve for t. Remember to convert all units to the appropriate SI units (meters, kilograms, and Newtons) before plugging them into the equation.

In your second problem, you are correct that it involves a moment of inertia. The moment of inertia is a measure of an object's resistance to rotational motion and is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

To solve this problem, you will first need to calculate the moment of inertia of the bucket. As you mentioned, for a cylindrical shape, the moment of inertia can be calculated by MR^2. Once you have this value, you can use it to calculate the minimum force required to raise the bucket. This can be done using the equation F=Iα, where I is the moment of inertia, α is the angular acceleration, and F is the force applied. In this case, α can be assumed to be constant since the bucket is being raised at a constant speed.

I hope this helps guide you in solving these torque problems. Remember to always pay attention to units and double-check your calculations. If you are still having trouble, don't hesitate to reach out for further