- #1
FilthyOtis
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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
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
