Tension in a string between two blocks

In summary, two masses M1 = 6.90 kg and M2 = 3.10 kg, connected by a string and on a frictionless surface, are being pulled by a force of 48.1 N at an angle of 31.3° from the horizontal. To calculate the tension T in the string, the forces acting on each block (gravity, normal force, and force from the pull) must be considered. The horizontal acceleration of the system can be found using the equation F=ma, using the correct value for F.
  • #36
Good. Now, what force(s) is/are acting on that block?
 
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  • #37
Brian T said:
Good. Now, what force(s) is/are acting on that block?
Tension is acting on that block.
 
  • #38
So, you know the force necessary to accelerate the block (~28) and you know there is only one force acting on the block, so...
 
  • #39
Brian T said:
So, you know the force necessary to accelerate the block (~28) and you know there is only one force acting on the block, so...
Is that the acceleration for the system then?
 
  • #40
You know the force needed to accelerate the block, and tension is the only force pulling it, therefore, the tension is equal to the force needed to accelerate it (~28 N)
 
  • #41
Brian T said:
You know the force needed to accelerate the block, and tension is the only force pulling it, therefore, the tension is equal to the force needed to accelerate it (~28 N)
So I have my answer?
 
  • #42
Yes. Another way to reach the answer is by looking at M2. This way is a bit more complicated since we consider more than one force but may be useful to look at for future preparedness.
We know that, since M2 is 3.9 kg, it should have a net force acting on it of:
F2net = ma
F2net = (3.9 kg)(4.10995 m/s^2)
F2net = 16.029 N.
Now the net force on block 2 should equal to the sum of the forces. The two forces are tension (left) and the pull (right). We have:
F2net = Fpull + T
F2net we just calculated is 16.029 N.
Fpull you previously calculated was 41.0995 N. Plugging it in:
16.029 N = 41.0995 N + T
Solve and get T ~ -28 (negative indicating left directed force. There is also an equal tension pulling to the right on block 1, which is the force you calculated).
 
  • #43
Brian T said:
Yes. Another way to reach the answer is by looking at M2:
We know that, since M2 is 3.9 kg, it should have a net force acting on it of:
F2net = ma
F2net = (3.9 kg)(4.10995 m/s^2)
F2net = 16.029 N.
Now the net force on block 2 should equal to the sum of the forces. The two forces are tension (left) and the pull (right). We have:
F2net = Fpull + T
F2net we just calculated is 16.029 N.
Fpull you previously calculated was 41.0995 N. Plugging it in:
16.029 N = 41.0995 N + T
Solve and get T ~ -28 (negative indicating left directed force. There is also an equal tension pulling to the right on block 1, which is the force you calculated).
Okay. Thank you so much for your help! I really appreciate you taking the time to help me!
 
  • #44
litz057 said:
Okay. Thank you so much for your help! I really appreciate you taking the time to help me!
No problem, glad I could help. Let me know if you have any more questions about that. :D
 
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  • #45
Thank you to everyone that helped me with this problem! I appreciate you taking the time to help me!
 
  • #46
Brian T said:
No problem, glad I could help. Let me know if you have any more questions about that. :D
I will! Thanks again!
 

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