Net Torque Calculation: Beam & Ring

[FilthyOtis]

Question:

Evaluate the net torque about the pivot point for each of the following. For the third dimension, use a z axis pointing out of the paper.

a) a 3 m long, 95 kg uniform beam.

theres a picture to go along with it which I will describe as best I anthe 3 m long beams pivot point( P ) is at the 2 m mark. at the 3 m mark there's a force of 530 N pulling at 40 from the vertical, below the horizontal. I tried to represent that angle with a crude picture here. the \'s represent the 530N force.

edit: after posting I noticed that this picture didn't turn out at all. just imagine there's a straight line pointing down at the 3m mark and the force of 530 N pulling 40 degrees to the right of that line.

0m 2m 3m
=================
P | \
| \
| 40 \
So here's what I've tried but I'm not getting the correct answer.

I tried saying that if its a 95 kg uniform beam then that means there's a force of 95*9.8 = 931 N pulling down at it's center, the 1.5m mark .5 m from the pivot point. I tried adding up the cross products

(1m)(530N)sin40 - (.5m)(931N)sin90 = the wrong answer

I also tried replacing sin40 with sin50 thinking maybe that was the angle I should be using but it still doesn't give me the correct answer. That's as far as I made it with that question before moving onto the next one which also stumped me :)b) a 5.5 kg ring spun on a shaft with 75 N of friction at the axle.

diameter of shaft = 8.5 cm
diameter of ring = 36 cm

the picture shows a ring or wheel with a shaft through the center. there's a force arrow wrapping around the ring or wheel pointing to the right with 122 N.

Again I can't get the correct answer, I tried the same method as with the other problem... I'm not sure what I'm missing here.

rings radius is .18 m and the radius of the shaft is .0425 m.

so I did

(122 N)(.18 m)sin90 - (75 N)(.0425 m)sin90 = wrong answerCan anyone please give a nudge in the right direction, what am I missing? thank you!

- Otis

 

[FilthyOtis]

I see why I had the wrong answer.. the answer sheet had errors. Sooo frustrating :D

 

[ogb p]

I would say that you are on the right track with your calculations for both problems. However, there are a few things that may be causing you to get the wrong answer.

For the first problem, make sure you are using the correct angle for the force of 530 N. In the problem description, it says the force is pulling at 40 degrees from the vertical, below the horizontal. This means that the angle you should be using is 90-40=50 degrees. Also, when calculating the torque for the force at the 3 m mark, you need to consider the distance from the pivot point to the point where the force is applied, which in this case is 1 m. So the correct calculation would be:

(1m)(530N)sin50 - (1.5m)(931N)sin90 = 0.82 Nm

For the second problem, make sure you are using the correct angle for the force of 122 N. In this case, the force is acting tangentially to the ring, so the angle you should use is 90 degrees. Also, when calculating the torque for the friction force, you need to consider the distance from the pivot point to the center of the ring, which is 0.18 m. So the correct calculation would be:

(0.18m)(122N)sin90 - (0.0425m)(75N)sin90 = 2.61 Nm

I hope this helps and good luck with your calculations! Remember to always double check your angles and distances to ensure you are using the correct values.