Physics Quiz 12

[Ether]

Originally Posted by Rachel Lim

Part 2:
I calculated the torque produced by gravity around point G, and the torque from tension around point E. I then subtracted the two values to find the overall torque.

You have to pick one point, and sum all the external torques around that one point, and set that sum equal to zero. Then solve.

If you pick the point carefully, the equation will have only one variable, so you can solve for it.

torque_gravity = (m*g)(distance_AG)(angle_AG to vertical)
= 15582*1.2*sin(20deg)
= 6395
torque_tension = (T)(distance_AE)(angle AE to ET)
= 5448*1.7*sin(140deg)
= 5953
So the overall torque around point A is 442 N*m in the counterclockwise direction.

Gravity and cable tension are not the only forces producing torques around point A.

What other external force(s) is(are) acting on the bucket+rollers+ore free body?


Mentors, Teachers, etc: Rachel is a very bright student. Let's give her (and any other students who may be following the thread) a chance to solve the problem.

[Ether]

To close out this thread, I'd like to invite any college students to jump in here and show how to solve part (b).

[Ether]

Originally Posted by Ether

To close out this thread, I'd like to invite any college students to jump in here and show how to solve part (b).

How about high-school physics teachers?

[mastachyra]

http://i.imgur.com/jYv6saa.jpg?1
http://i.imgur.com/SdCGmYc.jpg?1

I know this is pretty messy and the progress is kind of scattered, but here's my full FBD of the cart. Wheels A and D are contacting the rail: 16423 N at point A and 632 N at point D.

One of the things I think Rachel missed (though her work is very impressive) is the torque arm for the tension is 1.0 meter, which is found by extending the force vector backwards until a perpendicular line crosses through point A. So the torques are as follows:

torque_gravity around A = 6396 N*m
torque_tension around A = -5448 N*m
torque_(either C or D) = -948 N*m

The final point of contact must be point D so the force on the rail is downward.

Then its just a matter of finding the force on point A by setting the "vertical" forces equal to zero.

Sorry about the messy work.

[Ether]

Originally Posted by mastachyra

The final point of contact must be point D so the force on the rail is downward.

If D is contacting the rail then the force on the rail at D must be upward (perpendicular to the rail)

... and the force on the D roller is downward (perpendicular to the rail)

I know this is pretty messy and the progress is kind of scattered, but here's my full FBD of the cart. Wheels A and D are contacting the rail: 16423 N at point A and 632 N at point D.

Double-check that 632N number.

[mastachyra]

Originally Posted by Ether

... and the force on the D roller is downward (perpendicular to the rail)



Double-check that 632N number.

Whoops, thanks for clarifying the direction. Of course the force can only be down on the roller...

But I can't find the issue with 632N. The left over torque is -948 N*m and the torque arm from A to D is 1.5 m. no?

[Ether]

Originally Posted by mastachyra

But I can't find the issue with 632N. The left over torque is -948 N*m and the torque arm from A to D is 1.5 m. no?

Where did you get this from:

Originally Posted by mastachyra

the torque arm for the tension is 1.0 meter, which is found by extending the force vector backwards until a perpendicular line crosses through point A.

[mastachyra]

http://i.imgur.com/IU70nhu.jpg?1

OK, I see the error of my ways. I started from scratch (instead of trusting previous answers) and cleaned up my work to make deciphering a bit easier. I didn't bother with significant figures. New answer:

Part a) Tension on cable is 5454 N.
Part b) Force is 16082 N up on roller A and 291 N down on roller D.

I hope this is correct... mostly so we can get a new problem to work on!

[Ether]

Originally Posted by mastachyra

I hope this is correct... mostly so we can get a new problem to work on!

Close enough. See attachment.

[mastachyra]

Originally Posted by Ether

Where did you get this from:

Oh, you switched your question on me. I did things differently in my new answer, using the components instead of the actual tension on the cable. The 1.0 m torque arm was derived with the geometry that is circled and labeled TENSION TORQUE ARM. It's pretty crude, but I think it checks out. Here's clearer work on that part:
http://i.imgur.com/RCNAGre.jpg?1

[mastachyra]

Originally Posted by mastachyra

Oh, you switched your question on me. I did things differently in my new answer, using the components instead of the actual tension on the cable. The 1.0 m torque arm was derived with the geometry that is circled and labeled TENSION TORQUE ARM. It's pretty crude, but I think it checks out. Here's clearer work on that part:
http://i.imgur.com/RCNAGre.jpg?1

and i just caught the mistake in that too... the 8 degree angle should be 12. Which makes the 36 degree angle 40, and the torque arm, x = 1.1 m.