Atwood's machine problem - inclined plane

[don_anon25]

Thanks for the input!

 

[Physics Monkey]

Hi don,

I can't see the attachment for some reason, but if I understand the problem correctly, it looks like you did everything fine. You now simply need to solve for \theta using the equation
2 \sin{\theta} - 2 \mu_k \cos{\theta} - 1 = 0.
Unfortunately, the differential equation you have written is not equivalent to the equation you derived from Newton's Laws. This is because the differential has a solution that will be valid for all \theta (and will be exponential) while you are looking for a particular value of \theta. How do you solve for \theta then? Here is hint: try isolating \sin{\theta} and squaring both sides.

 

[Gokul43201]

Hi don,
I can't see the attachment for some reason,

All attachments go through a moderation queue and must be approved by a mentor beofre they are accessible. This one hasn't been approved just yet.

...but if I understand the problem correctly, it looks like you did everything fine. You now simply need to solve for \theta using the equation
2 \sin{\theta} - 2 \mu_k \cos{\theta} - 1 = 0.
Unfortunately, the differential equation you have written is not equivalent to the equation you derived from Newton's Laws. This is because the differential has a solution that will be valid for all \theta (and will be exponential) while you are looking for a particular value of \theta. How do you solve for \theta then? Here is hint: try isolating \sin{\theta} and squaring both sides.

...or write the cosine in terms of the sine, and then do the above.

There's one interesting detail with this problem though. It doesn't say which way the blocks are moving and so, there's no reason for the block on the incline to not be moving up the incline at a constant speed. This will give you another value of the incline angle.

And in fact, any angle between these two limiting cases will have the blocks moving at constant velocity. Think about how that can be true...

PS : This belongs in Introductory Physics.

 

[Physics Monkey]

Quite true, Gokul. Hopefully the down angle is sufficient for his problem.

However, I do think there is a slight mistake with your second point. The force of kinetic friction is independent of speed, it doesn't just match the forces the way static friction does. So if one were to replace \mu_k with \mu_s and then solve the "about to move down" and the "about to move up" cases, the block could sit at rest at any angle in between these two extreme cases owing to the fact that static friction will be whatever it takes to just keep the blocks at rest. However, if the block is initially moving, then there are only two angles, the up angle and the down angle, where the force of kinetic friction balances the other forces. What do you think?

BTW: Thanks for Homework Helper nod. What fun.

 

[Gokul43201]

Quite true, Gokul. Hopefully the down angle is sufficient for his problem.
However, I do think there is a slight mistake with your second point. The force of kinetic friction is independent of speed, it doesn't just match the forces the way static friction does. So if one were to replace \mu_k with \mu_s and then solve the "about to move down" and the "about to move up" cases, the block could sit at rest at any angle in between these two extreme cases owing to the fact that static friction will be whatever it takes to just keep the blocks at rest. However, if the block is initially moving, then there are only two angles, the up angle and the down angle, where the force of kinetic friction balances the other forces. What do you think?

I think it's one of those problems that turns out to reveal more neat stuff than the teacher expected. At angles between the two limiting cases, the blocks would, as you said above, decelerate from the initial velocity... and come to rest (rather than change direction, because at the instant that v=0 they'd experience static friction which will then take over). So, in the steady state, they'd have a constant velocity.

BTW: Thanks for Homework Helper nod.

You bet !

 

[Physics Monkey]

Great, glad to see we are in agreement.