In my 5 Steps to a 5 AP Physics prep book, page 70 is a review of tension problems, also known as many-body problems. I give eight different situations in which blocks are connected by ropes. The goal of each problem is to find the tension(s) in the rope(s), and the acceleration of the system.

The approach that I advocate is to draw a separate free body diagram for each block, then write Newton's second law separately for each block. The acceleration and tension(s) are solved for by adding the Newton's second law equations together.

Ruth Mickle, of Atlanta, noted yesterday that she gets a different answer to problem 6 than is printed in the book. I agree -- for whatever reason, the answer in my book is wrong. Below I give a thorough solution.

The problem shows three blocks connected by strings over a pulley, as shown at the top of the post. Given that

*m*is 1.0 kg, the question asks for the tensions and acceleration.Start by drawing three free body diagrams. Note that the two ropes will have two tensions; I'll label these

*T*1 and*T*2.The acceleration will be toward the heavier blocks. Thus, the mass

*m*will accelerate upward, and the other masses will accelerate downward. So when we write Newton's second law, for mass*m*we'll write "up forces - down forces =*ma.*" For the other masses, we'll write "down forces - up forces =*ma*." Always start Newton's second law in the direction of the acceleration.*T*2 -

*mg*=

*ma*2

*mg*+

*T*1 -

*T*2 = 2

*ma*4

*mg*-

*T*1 = 4

*ma*

Now, add 'em up. Note that the

*T*1s and the*T*2s will cancel in the addition:-

*mg*+ 2*mg*+ 4*mg*= 7*ma*Solving for

*a*, we get*a*= (5/7)*g*, or**7.1 m/s**^{2}.Now just plug back into the equations above to find that

**, and***T*2 = 17 N**.***T*1 = 11 N
I'm so glad I found this...I've been working on this problem for at least an hour and I wasn't getting the answer in the book! Thanks!!

ReplyDeleteI'm not a physics expert, but I think there are more errors than you realize...For example, on page 127 in the 08-09 book, the acceleration of a block on an inclined plane is written as = mu*g*cos(theta).

ReplyDeleteWhere did the "mgsin(theta)" go (from the original horizontal force summation? I keep getting that a = g(sin(theta)-mu*cos(theta))

If the answer in the book is right, can you let me know what I'm doing wrong? Thanks.

Rachel, that error has been corrected in the 2010-11 edition. For that 2010-11 edition, we had someone go through every question and solution thoroughly. You're right that there were (and are) errors, but many were rooted out unmercifully.

ReplyDeleteGCJ

I had forgotten how to do this kind of problem, so this was so incredibly helpful!!!! Thanks!!

ReplyDeletewhoa crazy, i looked this up on google because i was having trouble with this exact problem!! thank you so much :)

ReplyDeleteoh i just read that comment above - the answer for #6 in the 2010-2011 edition, the answer is a = 8.6 m/s^2, T1=19 N, and T2=9 N

ReplyDeleteThank you Master Jacobs, I have spent hours working this problem every which way and I kept getting your answer; not the books. Thanks for validating my work.

ReplyDeleteWell that's three hours of review time i'll never get back..

ReplyDeleteThanks for the heads up!

Derive an expression for tension & acc.

ReplyDeletewhy are the tensions different?

ReplyDeleteThe tensions are different because the top rope has to hold back both of the masses or a total of 6m, the bottom rope is only holding back 4m. This is consistent with the answer as the top rope holds 1.5 the mass of the bottom and so has 1.5(with rounding) the tension of the bottom rope

DeleteThanks...I spent hours for this

ReplyDeletethe sum is a little tricky and I had forgotten everything so it was a mind boost

ReplyDelete