Physics Quiz 12

[GeeTwo]

Originally Posted by DonRotolo

I couldn't find any of those frictionless rollers at McMaster, got a source for them?

I think I saw some last time I was at Theory Depot.

[Rachel Lim]

I attempted this yesterday, and I'm still not sure if I'm doing this correctly. I feel like I'm missing some step, but what I've done so far is below.

Assumptions / notes for part 1:
- The only thing that matters is the gravity component parallel to the rail, and the tension force parallel to the rail.
- The cart moves as a single body along the rail (it doesn't matter which part of the cart the rope is attached to)
- The 0.5 m/s is irrelevant, what does matter is that it's moving at a constant velocity (i.e. acceleration = net force = 0)

Part 1:
T = 5448 N

I'll post a picture of my working when I get a chance, but basically what I had was:
F_gravity = F_tension
m*g*sin(20deg) = T cos(12deg)
(All are the component parallel to the 20deg rail)

Assumptions / notes for part 2:
- The angle of the rope is less than the angle of the rail, so the cart is being pulled "into" the rail.
- This creates torque around point G and the cart is basically being twisted.

Part 2:
I'm not entirely sure how to do the calculations on this, but I'm guessing it results in rollers A and C being pulled "down" and B and D pulled "up."

[GeeTwo]

A hint for Rachel (and all),
For part 2, both the weight of the cart and the tension in the cable produce torques about any center of rotation convenient to calculating either of the roller forces.

[Ether]

Originally Posted by Rachel Lim

Assumptions / notes for part 1:
- The only thing that matters is the gravity component parallel to the ramp, and the tension force parallel to the ramp.

This is the first necessary insight.

- The cart moves as a single body along the ramp (it doesn't matter which part of the ramp the rope is attached to)

I think you meant the rail, the cable, and the bucket?

- The 0.5 m/s is irrelevant, what does matter is that it's moving at a constant velocity (i.e. acceleration = net force = 0)

Correct. This is the second necessary insight.

The 0.5 m/s was indeed a red herring. Part of solving physics problems is eliminating irrelevant information.

Part 1:
F_gravity = F_tension
m*g*sin(20deg) = T cos(12deg)
T = 5448 N

Correct!

[Ether]

Originally Posted by Rachel Lim

Part 2:
I'm not entirely sure how to do the calculations on this...

Do you want to continue working on it by yourself? Or shall I invite high-school teachers to chime in?

[Rachel Lim]

Originally Posted by Ether

Do you want to continue working on it by yourself? Or shall I invite high-school teachers to chime in?

Sorry, I had homework...what I have is below, but I welcome all other input.

I think I fixed the ramp / rail / cart thing in my previous post (I should really proofread more carefully).


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. For example, for point A:

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.

With this method, I got that all four rollers have torque in the counterclockwise direction, of magnitudes 442, 402, 24082, and 24092 Nm (A, B, C, D).

This is where I get stuck again, because I don't know how this correlates to figuring out which rollers make contact--is it the magnitude of the torque? The difference in the respective pairs?

[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!

[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