Newton's 2nd Law Problem

  • Thread starter wcase
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  • #1
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The problem is:
In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 31 kg, mB = 40 kg, and mC = 13 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.250 s (assuming it does not reach the pulley)?

Where block A is sitting on a table connected to a string that goes to a pulley at the end of the table where bock B and C are hanging.


I've read this question all over the internet and cannot interpret the results, I see whats being done, but I don't understand.

For the string between A and B, the tension is said to be mA*a, but why is there no influence from block B on the tension of the first string? shouldn't the tension be (mB+mC)*g since there would be no tension if the two blocks are pulling it down?

For block B, the net Force Fb=mB*a so 40a = Fb, and since the tension on the string between A and B is pulling it up and the tension on the string between B and C is pulling it down, 40a=40g-T1+T2.

For block C Fc=13a=13g+T2
so T2=13a-13g
and T2=40a-40g+T1

And that is where I am stuck, I do not know how to get T1
 

Answers and Replies

  • #2
Doc Al
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For the string between A and B, the tension is said to be mA*a, but why is there no influence from block B on the tension of the first string?
Why do you think there's no influence from block B?
shouldn't the tension be (mB+mC)*g since there would be no tension if the two blocks are pulling it down?
If the tension equaled that, then the blocks would be in equilibrium.

For block B, the net Force Fb=mB*a so 40a = Fb, and since the tension on the string between A and B is pulling it up and the tension on the string between B and C is pulling it down, 40a=40g-T1+T2.
Good.

For block C Fc=13a=13g+T2
Careful with signs. On C, the string tension acts up (opposite to gravity).

And that is where I am stuck, I do not know how to get T1
Write a force equation for block A.
 
  • #3
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With block A, the force from the string is pulling it, and the force is coming directly from the weight of block B and C, so I still don't understand why the weight pulling on the string has no effect on the tension of the string.
 
  • #4
Doc Al
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With block A, the force from the string is pulling it, and the force is coming directly from the weight of block B and C, so I still don't understand why the weight pulling on the string has no effect on the tension of the string.
What pulls block A is the tension in the string. The weight of block B and C pulls on B and C, not on A directly. Of course the weight of the hanging blocks affects the tension in the string, but it doesn't equal that tension.
 
  • #5
fgb
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Adding to what Doc Al said, the weight of blocks B and C is what drives the system to accelerate - when you write T1 = mA*a, for instance, the influence of blocks B and C is embedded in the value of the acceleration (different blocks B and C would result in a different acceleration).
 

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