Tension in a string between two blocks

AI Thread Summary
The discussion revolves around calculating the tension in a string connecting two masses on a frictionless surface, with a force applied at an angle. The correct approach involves determining the horizontal acceleration of the entire system using the net force and total mass. The tension in the string can then be derived from the forces acting on each mass, specifically considering the acceleration required to move mass M1. The participants clarify that gravitational forces do not affect the horizontal motion due to the frictionless condition, and they emphasize the importance of using the correct components of the applied force. Ultimately, the tension is found to be approximately 28 N, confirming the calculations and understanding of the forces involved.
litz057
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Homework Statement


Two masses M1 = 6.90 kg and M2 = 3.10 kg are on a frictionless surface, attached by a thin string. A force of 48.1 N pulls on M2 at an angle of 31.3° from the horizontal as shown in the figure. Calculate the tension T in the string.

Homework Equations


T=ma
a=g(m2/(m1+m2))

pullbox.gif


The Attempt at a Solution


a=g(m1/(m1+m2))
a=9.81m/s^2(3.10 kg /(6.90 kg + 3.10 kg))
a=9.81m/s^2(3.10 kg/10 kg)
a=9.81m/s^2(.31)
a=3.0411 m/s^2
T=ma
T=(6.90 kg)(3.0411 m/s^2)
T=20.98359 N
 
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So consider the force applied on the total mass. What should the masses accelerate? From there, isolate the system and see the tension needed to accelerate that isolated system by that amount
 
Brian T said:
So consider the force applied on the total mass. What should the masses accelerate? From there, isolate the system and see the tension needed to accelerate that isolated system by that amount
The forces that are applied on the mass are F (48.1 N), gravitational force, and the normal force. Did I find the acceleration incorrectly in my work above?
 
Hello Litz, welcome to PF :)

Your first relevant equation is OK, provided you mean M1 for m.

But your second one comes from elsewhere and doesn't fit here . In this exercise there is no friction with the surface, so the acceleration from gravity doesn't play a role.
Instead, consider a second equation just like the first one, but for M2. There are two forces working on M2 that are relevant for this exercise. Can you distinguish which two I mean ?
 
BvU said:
Hello Litz, welcome to PF :)

Your first relevant equation is OK, provided you mean M1 for m.

But your second one comes from elsewhere and doesn't fit here . In this exercise there is no friction with the surface, so the acceleration from gravity doesn't play a role.
Instead, consider a second equation just like the first one, but for M2. There are two forces working on M2 that are relevant for this exercise. Can you distinguish which two I mean ?
Is the static frictional force and F (force from the pull) what you are referring to? I have been struggling with determining forces and questions like the one I posted. My TA and professor have not been helpful when explaining the concepts to me.
 
You do not have to consider the weight since its frictionless. You just need to consider forces in the the direction of motion

What do you think is the horizontal acceleration of the whole system? Consider this before breaking it down to look at tension
 
Force from the pull is certainly relevant. The other one I meant has to do with T. There's a Newton law action = -reaction: if the connecting wire pulls to the right on M1 with a force T, then it pulls with an equal but opposite force -T to the left on M2.

The static frictional force is zero on a frictionless surface. The two other forces that work on each of the two blocks are gravity (M1g and M2g, respectively) working downwards, and the normal forces the surface exerts upwards. These simply cancel (the blocks don't accelerate in the vertical direction, so in that diretiction the sum of forces is zero (*) ) .

So for block 1 we have something with a T and an a. These also appear when we write the equation for block 2. And there F has to show up too, but not the full 48.1 N. Do you know whereI'm trying to go ?
 
BvU said:
Force from the pull is certainly relevant. The other one I meant has to do with T. There's a Newton law action = -reaction: if the connecting wire pulls to the right on M1 with a force T, then it pulls with an equal but opposite force -T to the left on M2.

The static frictional force is zero on a frictionless surface. The two other forces that work on each of the two blocks are gravity (M1g and M2g, respectively) working downwards, and the normal forces the surface exerts upwards. These simply cancel (the blocks don't accelerate in the vertical direction, so in that diretiction the sum of forces is zero (*) ) .

So for block 1 we have something with a T and an a. These also appear when we write the equation for block 2. And there F has to show up too, but not the full 48.1 N. Do you know where I'm trying to go ?
Not really. I think I might have an idea, but I do not think that it is correct. Since the full 48.1 does not show up, something needs to be subtracted. Am I supposed to find the T=m1a, then subtract that from the equation for block 2?
 
Brian T said:
You do not have to consider the weight since its frictionless. You just need to consider forces in the the direction of motion

What do you think is the horizontal acceleration of the whole system? Consider this before breaking it down to look at tension

I am not sure what is the horizontal acceleration of the whole system. Do I use F=ma to find the acceleration of the whole system?
 
  • #10
The net force on the system will accelerate the mass of the system by that amount proportionally. Other force pairs are internal to the system and do not change the net force acting on our system, only how the forces act. So, yes
 
  • #11
litz057 said:
I am not sure what is the horizontal acceleration of the whole system. Do I use F=ma to find the acceleration of the whole system?
Yes, as long as you use the correct value for F in that equation.
 
  • #12
haruspex said:
Yes, as long as you use the correct value for F in that equation.
I was told earlier that I cannot use 48.1 N as F, so how do I figure out what F is for this problem?
 
  • #13
Read my post earlier about considering force in the direction of motion
 
  • #14
litz057 said:
I was told earlier that I cannot use 48.1 N as F, so how do I figure out what F is for this problem?
You wrote:
litz057 said:
what is the horizontal acceleration of the whole system. Do I use F=ma to find the acceleration of the whole system?
So clearly we are discussing horizontal acceleration. What direction of force produces horizontal acceleration?
 
  • #15
haruspex said:
You wrote:

So clearly we are discussing horizontal acceleration. What direction of force produces horizontal acceleration?
The right?
 
  • #16
You're given a force vector and an angle. Break it down :)
 
  • #17
Brian T said:
Read my post earlier about considering force in the direction of motion
The force would have to be to the right because if it was to the left the blocks would not accelerate...?
 
  • #18
Brian T said:
You're given a force vector and an angle. Break it down :)
Do I use Fcos(theta) (48.1*cos(31.3))?
 
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  • #19
Mhmm. From there, what is the net acceleration? After this, try to think about the forces that are acting on both blocks, the magnitude of the forces that should act on both blocks. If you think about this, you can make your way toward the tension of the rope.
 
  • #20
Brian T said:
Mhmm. From there, what is the net acceleration?
Is the acceleration 48.1*cos(31.3)=41.0995?
 
  • #21
litz057 said:
Is the acceleration 48.1*cos(31.3)=41.0995?
That is the force in the direction of the acceleration, not the acceleration itself.
 
  • #22
Brian T said:
Mhmm. From there, what is the net acceleration? After this, try to think about the forces that are acting on both blocks, the magnitude of the forces that should act on both blocks. If you think about this, you can make your way toward the tension of the rope.
Are the forces that you are talking about T (tension) and F (force from the pull)?
 
  • #23
Brian T said:
That is the force in the direction of the acceleration, not the acceleration itself.
If it is the force in the direction of the acceleration, then is it the Fnet?
 
  • #24
That's correct. The goal here is to try to figure out what the blocks should be accelerating at, and then determine the partial force of the tension necessary in order to achieve that acceleration.

What I said in the last post, all I'm saying is you computed the force relevant, not the acceleration relevant. How do we get from force to acceleration?
 
  • #25
Brian T said:
That's correct. The goal here is to try to figure out what the blocks should be accelerating at, and then determine the partial force of the tension necessary in order to achieve that acceleration.

What I said in the last post, all I'm saying is you computed the force relevant, not the acceleration relevant. How do we get from force to acceleration?
So how do I relate the force in the direction of the acceleration to figure out the partial force of the tension (to find the acceleration)?
 
  • #26
Use Newton's law to figure out the acceleration of the whole system, now that you have F and you already know m.

This is the more general question to answer before looking at the specific question of tension.
 
  • #27
Brian T said:
Use Newton's law to figure out the acceleration of the whole system, now that you have F and you already know m.

This is the more general question to answer before looking at the specific question of tension.
I am confused about what mass to use. Do I use M1 or M2?
 
  • #28
If you're consider the force acting on the whole system, and want to know the acceleration of the whole system, you would need the mass of?
 
  • #29
Brian T said:
If you're consider the force acting on the whole system, and want to know the acceleration of the whole system, you would need the mass of?
M2
 
  • #30
Use the mass of the whole system to find the acceleration of the whole system

I.e. M1 + M2
 
  • #31
Brian T said:
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
 
  • #32
Yes, so a =?
 
  • #33
Brian T said:
Yes, so a =?
F=ma
41.0995=10a
41.0995/10=a
4.10995=a
 
  • #34
So, you have the acceleration of the system. Now, look at block m1. What is the force needed to accelerate it at 4.10995?
 
  • #35
Brian T said:
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
 
  • #36
Good. Now, what force(s) is/are acting on that block?
 
  • #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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