Tension in a string between two blocks

[litz057]

Use the mass of the whole system to find the acceleration of the whole system

So I use M₁ + M₂ = 6.9 +3.1 = 10 kg

 

[Brian T]

Yes, so a =?

 

[litz057]

Yes, so a =?

F=ma
41.0995=10a
41.0995/10=a
4.10995=a

 

[Brian T]

So, you have the acceleration of the system. Now, look at block m₁. What is the force needed to accelerate it at 4.10995?

 

[litz057]

So, you have the acceleration of the system. Now, look at block m₁. What is the force needed to accelerate it at 4.10995?

F=ma
F=6.9kg*4.10995m/s^2
F= 28.3586 N

 

[Brian T]

Good. Now, what force(s) is/are acting on that block?

 

[litz057]

Good. Now, what force(s) is/are acting on that block?

Tension is acting on that block.

 

[Brian T]

So, you know the force necessary to accelerate the block (~28) and you know there is only one force acting on the block, so...

 

[litz057]

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?

 

[Brian T]

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)

 

[litz057]

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?

 

[Brian T]

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:
F_2net = ma
F_2net = (3.9 kg)(4.10995 m/s^2)
F_2net = 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:
F_2net = F_pull + T
F_2net we just calculated is 16.029 N.
F_pull 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).

 

[litz057]

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:
F_2net = ma
F_2net = (3.9 kg)(4.10995 m/s^2)
F_2net = 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:
F_2net = F_pull + T
F_2net we just calculated is 16.029 N.
F_pull 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!

 

[Brian T]

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

 

[litz057]

Thank you to everyone that helped me with this problem! I appreciate you taking the time to help me!

 

[litz057]

No problem, glad I could help. Let me know if you have any more questions about that. :D

I will! Thanks again!