Three blocks pulled by a force

In summary, the two masses are accelerating and the tension in the string connecting them is not equal. The tension is due to the force of 20 N and the inertia of the masses.
  • #1
rudransh verma
Gold Member
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Homework Statement
Three blocks of mass 10, 6,4 kgs are placed on friction less surface. Force of 40N pulls the system. Calculate T.
Relevant Equations
##F=ma##
$$a= -40/(10+6+4)$$
$$a=-2 m/s^2$$
Taking one mass of 10 kg.
$$T-40=10(-2)$$
$$T=20 N$$
This is correct.
But if I make the eqn of the system then
$$-40+T-T+T-T=20(-2)$$
I have also drawn the diagram. It looks like the second body m2 is subject to no force. But it’s accelerating. How?
 

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  • #2
Your assumption that the tension in all strings is the same is wrong.
 
  • #3
rudransh verma said:
I have also drawn the diagram. It looks like the second body m2 is subject to no force. But it’s accelerating. How?
If you say that the mass is accelerating and the diagram you drew shows that it isn't, then either you are wrong or the diagram you drew is wrong. Which do you think is the case?
 
  • #4
Orodruin said:
Your assumption that the tension in all strings is the same is wrong.
First of all there is no friction. The system will accelerate.
Does tension depend on mass?
 
  • #5
rudransh verma said:
First of all there is no friction.
So what? My statement did not assume that there was.
 
  • #6
Orodruin said:
So what? My statement did not assume that there was.
rudransh verma said:
Does tension depend on mass?
 
  • #7
rudransh verma said:
Does tension depend on mass?
It does. If you have a train of ##N## masses pulled by a constant external force, the tension between mass ##m_k## and ##m_{k+1}## will depend on all the masses following the ##k##th mass starting with ##m_{k+1}##.
 
  • #8
rudransh verma said:
Does tension depend on mass?
This statement is nonsensical. What mass? Under what conditions? You need to specify.
 
  • #9
Orodruin said:
Your assumption that the tension in all strings is the same is wrong.
@rudransh verma :
The statement made by @Orodruin seems clear enough. But maybe it is not clear enough.

The tension in the string connecting block1 with block2 is not equal to the tension in the string connecting block2 with block3 .
 
  • #10
kuruman said:
It does. If you have a train of N masses pulled by a constant external force, the tension between mass mk and mk+1 will depend on all the masses following the kth mass starting with mk+1.
So I calculated T1= 20N
And T2= 8N.
And this tension is due to force 40N and inertia of the masses since there is no friction?
Isn’t tension due to all masses, applied force and friction ?
 

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  • #11
rudransh verma said:
Why not also mk mass?
Because mass ##m_k## cannot exert a force on itself, only on the masses that it's pulling. In the system that you have there could be a fourth 100 kg or 1 kg or any mass exerting a force of 40 N on the front mass ##m_1## and it wouldn't make a difference.
rudransh verma said:
So I calculated T1= 20N
And T2= 8N.
These numbers are correct.
rudransh verma said:
And this tension is due to force 40N and inertia of the masses since there is no friction?
Yes, in a sense. The three masses subdivide the 40 N net force in the following manner
$$Ma=(m_1+m_2+m_3)a=m_1a+m_2a+m_3a=(20+12+8)~\text{N}=40~\text{N}=F_{\text{Net}}$$The difference between the two tensions acting on the middle mass must be equal to ##m_2a##. That's what you missed when you started solving this problem.
 
  • #12
kuruman said:
Because mass mk cannot exert a force on itself, only on the masses that it's pulling. In the system that you have there could be a fourth 100 kg or 1 kg or any mass exerting a force of 40 N on the front mass m1 and it wouldn't make a difference.
Yeah! So it depends on the applied force. Since applied force is in left of 40N ,tension in every string is due to m2 and m3 for T1 and m3 for T2.
kuruman said:
manner
Ma=(m1+m2+m3)a=m1a+m2a+m3a=(20+12+8) N=40 N=FNetThe difference between the two tensions acting on the middle mass must be equal to m2a. That's what you missed when you started solving this problem.
20N is tension in first string and 8 in second. What is 12N. You say it’s the difference in tensions on mass m2.
So this is not the tension in string? I don’t understand! It should be in the strings.
 
  • #13
rudransh verma said:
20N is tension in first string and 8 in second. What is 12N. You say it’s the difference in tensions on mass m2.
So this is not the tension in string? I don’t understand! It should be in the strings.
The free body diagram for m₂ [edit - ignoring the vertical forces which cancel] is:

T₁←[m₂]→T₂

From this free body diagram, you can see the net force on m₂ is T₂ – T₁
##F_{net,2} = T₂ – T₁ = 8 – 20 = -12N##

That means m₂’s acceleration is
##a = \frac {F_{net,2}} {m ₂} = \frac {-12} {6} = --2m/s^2## as expected.

So it all makes sense..
 
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  • #14
rudransh verma said:
20N is tension in first string and 8 in second. What is 12N. You say it’s the difference in tensions on mass m2.
So this is not the tension in string? I don’t understand! It should be in the strings.
You are tripping over your own two feet.

Mass m1 is pulling to the left on m2 with a force of 20 N. This is T1.
Mass m3 is pulling to the right on m2 with a force of 8 N. This is T2.
To the left is negative and to the right is positive, so T1 = -20 N and T2 = +8 N.
The net force on m2 is the sum:
Fnet = T1 + T2 = -20 N + 8 N = -12 N
Mass times acceleration for m2 is m2*a = 6 (kg)*(-2) (m/s2) = -12 N.

See how Newton's second law works for m2?
 
  • #15
rudransh verma said:
Yeah! So it depends on the applied force. Since applied force is in left of 40N ,tension in every string is due to m2 and m3 for T1 and m3 for T2.

20N is tension in first string and 8 in second. What is 12N. You say it’s the difference in tensions on mass m2.
So this is not the tension in string? I don’t understand! It should be in the strings.
You will have better success solving these problems if you desist from trying to take a holistic view. Yes, there are times when that can be a faster route, but it seems to mislead you too often.

Separate it out into a free body diagram for each rigid body. Relate them only by the forces that act between directly connected bodies and the motions they must share. Keep in mind that each body only "knows" about the forces that act directly on it. Likewise, the action and reaction between two connected bodies don't "know" anything about other action/reaction pairs in the system.

So here you have ##T_{12}## acting between ##m1, m_2##; ##T_{23}## acting between ##m3, m_2##; the force 40N acts only on ##m_1##; and all three having the same acceleration, ##a## in the direction of the applied force.
From that you can write a ##\Sigma F=ma## equation for each block, and you have three unknowns. Solve.
 
  • #16
haruspex said:
You will have better success solving these problems if you desist from trying to take a holistic view. Yes, there are times when that can be a faster route, but it seems to mislead you too often.

Separate it out into a free body diagram for each rigid body. Relate them only by the forces that act between directly connected bodies and the motions they must share. Keep in mind that each body only "knows" about the forces that act directly on it. Likewise, the action and reaction between two connected bodies don't "know" anything about other action/reaction pairs in the system.

So here you have ##T_{12}## acting between ##m1, m_2##; ##T_{23}## acting between ##m3, m_2##; the force 40N acts only on ##m_1##; and all three having the same acceleration, ##a## in the direction of the applied force.
From that you can write a ##\Sigma F=ma## equation for each block, and you have three unknowns. Solve.
Yeah! I completely forget often.
 

Related to Three blocks pulled by a force

What is the concept of "three blocks pulled by a force"?

The concept of "three blocks pulled by a force" refers to a physics problem in which three blocks are connected by strings and pulled by a force in the same direction. This problem is often used to illustrate concepts such as Newton's laws of motion and forces in equilibrium.

How do you calculate the acceleration of the blocks in this scenario?

To calculate the acceleration of the blocks, you can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this scenario, the net force is the force pulling the blocks, and the mass is the combined mass of the three blocks. By dividing the net force by the combined mass, you can determine the acceleration of the blocks.

What is the difference between static and kinetic friction in this scenario?

In this scenario, static friction refers to the force that prevents the blocks from sliding against each other before the force is applied. Kinetic friction, on the other hand, refers to the force that opposes the motion of the blocks as they begin to slide against each other. Both types of friction are dependent on the materials and surfaces involved.

How does the force affect the tension in the strings connecting the blocks?

The force pulling the blocks will cause tension in the strings connecting them. As the force increases, the tension in the strings will also increase. This is because the tension in the strings must be equal to the force pulling the blocks in order to keep them in equilibrium.

What happens to the acceleration if the mass of one of the blocks is increased?

If the mass of one of the blocks is increased, the acceleration of the blocks will decrease. This is because the net force acting on the blocks remains the same, but the combined mass of the blocks has increased. According to Newton's second law of motion, a larger mass will result in a smaller acceleration.

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