Three blocks pulled by a force

[rudransh verma]

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?

 

[Orodruin]

Your assumption that the tension in all strings is the same is wrong.

 

[kuruman]

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?

 

[rudransh verma]

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?

 

[Orodruin]

First of all there is no friction.

So what? My statement did not assume that there was.

 

[rudransh verma]

So what? My statement did not assume that there was.

Does tension depend on mass?

 

[kuruman]

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}##.

 

[Orodruin]

Does tension depend on mass?

This statement is nonsensical. What mass? Under what conditions? You need to specify.

 

[SammyS]

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 .

 

[rudransh verma]

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 ?

 

[kuruman]

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.

So I calculated T1= 20N
And T2= 8N.

These numbers are correct.

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.

 

[rudransh verma]

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.

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.

 

[Steve4Physics]

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..

 

[kuruman]

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:
F_net = 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?

 

[haruspex]

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.

 

[rudransh verma]

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.