How Does Tension Vary in a Multi-Block Pulley System?

In summary, the problem involves three connected ballot boxes with different masses. When released from rest, the tension in the cord connecting B and C can be found using the forces on block B and C. The distance block A moves in 0.250 s can be determined, but the tension in the cord connecting A and B remains unknown. The tension in the string is influenced by the weights of blocks B and C, but it does not directly equal the weight of the hanging blocks.
  • #1
wcase
9
0
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 what's 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
 
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  • #2
wcase said:
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
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
wcase said:
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
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).
 

Related to How Does Tension Vary in a Multi-Block Pulley System?

What is Newton's 2nd Law Problem?

Newton's 2nd Law Problem, also known as Newton's Second Law of Motion, is a fundamental principle in physics that explains the relationship between an object's mass, acceleration, and the force acting on it.

What is the formula for Newton's 2nd Law Problem?

The formula for Newton's 2nd Law Problem is F = ma, where F represents the force acting on an object, m represents the mass of the object, and a represents the acceleration of the object.

How is Newton's 2nd Law Problem applied in real life?

Newton's 2nd Law Problem can be applied in various real-life situations, such as calculating the force required to push or pull an object, determining the acceleration of a moving car, or understanding the effects of gravity on a falling object.

What are the units of measurement for Newton's 2nd Law Problem?

The units of measurement for Newton's 2nd Law Problem are Newtons (N) for force, kilograms (kg) for mass, and meters per second squared (m/s²) for acceleration.

What are some common misconceptions about Newton's 2nd Law Problem?

One common misconception is that an object's mass and acceleration are directly proportional. In fact, the mass of an object affects its acceleration, but it is not the only factor. Another misconception is that Newton's 2nd Law Problem only applies to objects at rest or in constant motion, when in reality it can also be applied to objects undergoing changes in motion.

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