Can I see the worked solution for 11d? Thank you.

John, I'm emailing it to you now.  John

Hiya! Can I get some help on #3 (worked out solution?).  I do not understand the diagram so I am struggling to get started.  Thanks!

Michelle, I'll send it to your email.  John

Hello, would it be possible for me to also get worked out solutions to this problem set? Thanks!

Jorge, we don't send out the complete set - but if you have questions on some of them, I can send them to you.  John

I am in the same boat with Michelle, up above. I can not get the final answers you got for #3, thanks in advance. 👍

Jorge, sending it now.  John

Hi, could I get solutions to #14 and #3?  For 14, I do not see why it is M+m and not just m on c/d.  Thank you!

Rich, I'm sending the solutions now.  For #14, we added the two simultaneous equations - the first one describing the forces on m and the second one describing the forces on M. The acceleration that M feels is a combination of the rope accelerating down (a sub b) and M accelerating down the rope at (a). When the two equations are added, that gives us an Ma that adds to a sub b, resulting in the M + m term.   John

Hello, could I get solutions for #3, #6 and #7?  Thanks!

Jane, I'll email them to you.  John

Hi, can I get solutions for #3, #6,d, and #7c? Thank you!

They John,  I am struggling with answering 2c and #3.

BD

Brandon, I sent you my solution for #3.

As for 2c, we have the top block, a1, accelerating to the right as it is being pulled by the force, F. It has a friction force acting to the left due to block a2. That same friction force is pulling the bottom block, a2, to the right (Newton's Third Law).

Both accelerations were found relative to the horizontal surface.

So, in the reference frame of block 2, the acceleration of block 1 is a1 minus a2. That's what we need to calculate how long it would take to cover the distance L.

John

For question #14, do you have the block accelerating upwards in your answer?  I had it accelerating downwards and have a - off from your answer.  Also, I feel that the problem really should not have the ring accelerating WITH RESPECT TO THE ROPE.  That is a very subtle difference.

Michael, for part c (I think that's what you're referring too), I set it up so the block is accelerating in a clockwise direction.  Just chose it randomly. But, when we get the solution, if we let a = g, we get the block accelerating in a counter clockwise direction.  And, why did you feel that the ring should not be referenced as accelerating with respect to the rope?  I want to ensure I'm looking at this problem correctly and answering you properly.  John

Would it be possible to get the worked-out solutions to these problems?  Thanks

Daniel, at this time, we don't make them all available, but will send individual ones as you request.

John

For question 2. When finding the acceleration of block m1 relative to the horizontal surface (m2), I got a1 =(F-mum1g)/m1 which is given in the answer key. For part C, why do we have to subtract the accelerations to find the acceleration relative to block m2? Isn't that what we found in part B? Thank you!

Katherine, that would be the acceleration relative to the ground. Since the bottom block is also moving to the right, the relative acceleration between the two blocks requires the subtraction. The bottom block is "running away" from the top block, so it takes longer for the top block to slide off the end.  John

Would I be able to have the worked out solution for problems 3,6, and 7. Thank you.

Sabrina,  Here are problems 3 and 7. I'll get #6 to you tomorrow.  John

Sabrina,  Here are problems 3 and 7. I'll get #6 to you tomorrow.  John

Thank you John. I fell into the trap of not reading the question completely.

We all fall into that trap!  John