Physics pulley question

[dnovikov]

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A little confused about how TBR explains their answer to discrete question #35 from physics chapter 2 in the TBR book. If someone can clarify it that would be awesome.

The question has a diagram which I can't really put on here but it goes like this:

"What is the net acceleration for the mass on the right side of the Atwood machine shown below?"

Its found on page 116 in the TBR Physics 1 book.

 

[MedPR]

A little confused about how TBR explains their answer to discrete question #35 from physics chapter 2 in the TBR book. If someone can clarify it that would be awesome.

The question has a diagram which I can't really put on here but it goes like this:

"What is the net acceleration for the mass on the right side of the Atwood machine shown below?"

Its found on page 116 in the TBR Physics 1 book.

I remember being confused about this one.

This cleared it right up for me.

http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

 

[chiddler]

can i get a drawing, please? i'm not too confident with pulley's and this would be good practice for me if it was confusing

 

[IH8ColdWeath3r]

if you have TPR, they explain this really well in the 2 accelerating bodies section. Basically though, an easy way to do pulley problems like this (or problems like when you have a block on a flat surface tied by a rope to a small pull that is attached to a Larger block that is free falling) is to mentally set up to free body diagrams, determine the direction of the system, and cancel out the tensions.


Ex. Take the case of an atwood machine with two blocks attached, a smaller block on the left designated m and a larger block on the right designated M. What is the acceleration of the system.

m= 5 kg
M= 10 kg

In order to solve for a, set up TWO free body diagrams, one for each mass. Then, DESIGNATE A DIRECTION depending on the motion of the box. In this case, the 10 kg box is heavier so it will move down, and the 5 kg box is lighter so it will move up. ***

**it is easiest to assign the direction of the motion of the object as POSITIVE. SO....

FBD for M should show the large box on the right, with gravity acting on it to pull it down (designated as positive direction). Since tension in the rope is in the opposite direction, that is designated as the negative direction (so we have NEGATIVE tension for the FBD of M)

Ma = W - T where W=Mg

FBD for m should show the small box on the left. Now, since the larger mass on the right is moving downward, the smaller box on the right must naturally move upward. Since this box's motion is upward, we will designate the positive as the upward direction. Thus, the tension becomes positive. Now, since the only other force on the small box is it's weight, which OPPOSES the direction of the motion of the box (thus well assign it negative) we need to put that in our equation.

ma = T- w where w=mg

Now, if you go back and read carefully, in the FBD for M, tension was designated as NEGATIVE (since it was in the opposite the direction of the motion of the larger box) and in the FBD for m, tension was designated as POSITIVE, since it was in the same direction as the motion of the smaller box). THUS, TENSIONS CANCEL. THIS IS IMPERATIVE.

Now, with the tensions canceling, we can look at the two FBD's and add them together. Remember the direction of the system and the directions that we assigned. Well get

Ma + ma = W - w
a (M+m) = W- w

a= (W-w)/(M+m)
a= (100N - 50N)/(10kg+5kg) = 3.33 m/s^2

you can take the acceleration and plug it back into either one of the equations and find the tension in the rope. Also remember that tension is constant throughout the rope! If your having trouble, go back and draw the problem and go through each step that I explained. I know it sounds confusing but I tried my best to explain. If you have the TPR book it is explained at the end of the first mechanics section I believe.

 

[chiddler]

Thank you very much for the helpful post!

 

[drechie]

this is incredibly helpful, but how does one solve this question using intuition. TBR's solution wasn't that clear and i didn't understand their rational. I didn't understand how they could cross out B and select A (i understood how they crossed out C and D).

This is a repeat of Example 2.3a. Hopefully you recall from that example that attempting to solve the question by traditional means (mathematics and the equating of forces) is not the most time-efficient method nor necessary on a multiple-choice exam. Plugging in extreme values and seeing which equation matches the predicted results is the ideal route to finding the best answer. Choice Cis the first to be eliminated, because the kg unit doesn't cancel out. Ifwe consider the case where mi is greater than m2, then we predict by observation that the system will accelerate downward on the left side. The magnitude of a should increase as mj increases and m2 decreases. Choice D definitely doesn't fit this observation, so it is eliminated. If m2 is greater than mi, then we would expect the mass on the left side to accelerate upwards, so the sign of a has to be different when m2 >mi than it was when m\ > m2. Only choice Afits this restriction. The

best answer is choice A.