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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 T1 and T2.
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.
T2 - mg = ma 2mg + T1 - T2 = 2ma 4mg - T1 = 4ma
Now, add 'em up. Note that the T1s and the T2s will cancel in the addition:
-mg + 2mg + 4mg = 7ma
Solving for a, we get a = (5/7)g, or 7.1 m/s2.
Now just plug back into the equations above to find that T2 = 17 N, and T1 = 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!!
ReplyDeleteThank u very much..
DeleteI'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
ReplyDeleteCan't we just add up 4m and 2m?
ReplyDeleteOn solving I got the desired answer
To get the acceleration of the system, yes - you can treat the 2m and 4m as a single object of mass 6m. There are multiple ways of approaching the problem. The one I suggest in the post is merely the one I advocate students to use... because the general method will always work, and is easily understood. As long as you can explain what you did, and as long as your approach is physically valid, go for it!
ReplyDeleteThanks....
ReplyDeleteamazing...
ReplyDeleteThanks a lot..
DeleteThank you
ReplyDeleteI have been trying it since an hour. Well, thanks!
ReplyDeleteReally helpful.Thanks yah
ReplyDeleteHow to calculate tension on pulley
ReplyDeleteIn the equation for m block, it is written that it should be "up forces-down forces", then shouldn't it be
ReplyDeletema+T2=mg instead of mg+T2=ma? If not, then could you please explain why? Thank you so much for this concept. It really helped
Unknown, "ma" is the expression for the *unbalanced* force - it's not a force acting on a block. The forces acting on the block of mass m are T2 upward, and mg downward. Since the acceleration is upward, the forces are unbalanced upward, so the unbalance force is T2-mg. The unbalanced force is set equal to ma, giving T2-mg=ma.
ReplyDelete