Crane Lifting 186 kg Crate: Torque Calculation

[PhyzicsOfHockey]

Homework Statement

The crane shown in the drawing is lifting a 186-kg crate upward with an acceleration of 2.0 m/s2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

Homework Equations

torque= I*angular acceleration
I= mr^2
angular accel= accel/r

The Attempt at a Solution

I tried everything that i know

angular accel= 2/.76 =2.6316 rad/s^2
I=150*.76^2= 86.64
torque= 86.64*2.6316= 228 N*m

But this answer is wrong. I tried negitive and that's wrong. What am I doing wrong! lol.

 

[anathema]

Thank you for your question. After reviewing your attempt at a solution, it seems that you have the correct equations and concepts in mind, but there are a few small errors in your calculations.

First, for the moment of inertia (I), you need to consider the mass of the pulley (130 kg) as well. So the correct calculation would be I = (150 + 130) * 0.76^2 = 191.52 kg*m^2.

Secondly, for the angular acceleration, you need to use the total mass of the system (186 kg + 130 kg + 150 kg = 466 kg) and divide by the total radius (0.76 m). So the correct calculation would be angular accel = 2 / (466 / 0.76) = 0.0032 rad/s^2.

Finally, using the correct values in the torque equation, we get torque = 191.52 * 0.0032 = 0.613 N*m. This is the magnitude of the torque being applied by the engine to the drum in order to lift the crate.

I hope this helps clarify any confusion and leads you to the correct answer. Keep up the good work in your science studies!

 

[m2003]

As a scientist, it is important to carefully analyze all of the given information and equations in order to determine the correct solution. In this case, the torque calculation should take into account the torque from both the drum and the pulley, as both are involved in the lifting process.

The torque from the pulley can be calculated using the equation torque = radius * force. In this case, the force is equal to the weight of the crate (186 kg * 9.8 m/s^2 = 1822.8 N), and the radius is equal to half the diameter of the pulley (0.38 m). Therefore, the torque from the pulley is 691.94 N*m.

The torque from the drum can be calculated using the equation torque = moment of inertia * angular acceleration. The moment of inertia for a hollow cylinder is equal to 1/2 * mass * radius^2, so in this case it is equal to 1/2 * 150 kg * (0.76 m)^2 = 57.3 kg*m^2. The angular acceleration is still 2.6316 rad/s^2. Therefore, the torque from the drum is 150.5 N*m.

To find the total torque, we can add the torques from the pulley and the drum together: 691.94 N*m + 150.5 N*m = 842.44 N*m. Therefore, the magnitude of the torque applied by the engine to lift the crate is 842.44 N*m.