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This was a problem in an examination paper that I was solving:

Three blocks of masses $m_1$, $m_2$ and $m_3$ are connected to a massless string on a frictionless table as shown in the figure. They are pulled with a force $F = 40 N$ . If $m_1 = 10 kg$ , $m_2 = 6 kg$ , and $m_3 = 4 kg$ , then tension $T_2$ will be

(a) $10 N$

(b) $20 N$

(c) $32 N$

(d) $40 N$

Question diagram

I got to a system of equations, but I feel I missed something. The forces on $m_1$ are $F = 40 N$ in the right direction and tension $T_2$ in the left. The system is, of course, accelerating. So my first equation is $F - T_2 = m_1a$ .

The problem arises when I get to the second block. I'm really confused as to what forces are acting on the second block, and in which direction. I feel that is where I went wrong. The first time I did this question I actually made a silly mistake and considered a force that wasn't acting on the second block, but was in fact being applied by the second block (tension $T_2$ in the right direction). The diagram doesn't help much either. What's with the direction of $T_1$ and $T_2$ given in the diagram?

I need a little help here. Free body diagrams for the three blocks would help greatly.

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    $\begingroup$ All 3 blocks experience the same acceleration so. You can treat them as a single block to find the acceleration produced by F. $\endgroup$ – C. Towne Springer Apr 11 '18 at 6:27
  • $\begingroup$ Tension always pulls $\endgroup$ – SmarthBansal Apr 11 '18 at 12:08
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When summing the forces on each mass, you only take into account the forces that are acting directly on them. So, the 40N doesn't get summed for the second mass because that mass only has T1 pulling it backward and T2 pulling it forward. Don't forget to sum the forces on the last mass and start combining the equations to solve. enter image description here

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The forces shown on your diagram might well be there to help you but also thay seem to have confused you.
If you are going to use Newton's second law then first define the system you are considering and then draw the appropriate free body diagram.

The free body diagrams showing only the horizontal forces for the three masses and the two strings (each regarded as a separate system) are shewn below.

enter image description here

Forces a and b (force on mass $m_3$ due to left hand string and force on left hand string due to mass $m_3$), c and d, e and f, g and h are all Nweton's third law pairs (equal in magnitude and opposite in direction.

Forces b and c, f and g represent the forces on the two strings which, the the strings are assumed massless, are equal in magnitude and opposite in direction so the net force on each string is zero.

Because the strings are also assumed to be inextensible the acceleration of each of the masses must be the same.

However you can define your system in different ways.
If you take mass $m_3$ and mass $m_2$ as one system then the external force which acts on that system is force e of magnitude $T_2$ towards the right.
All the other forces a, b, c, d are internal forces and so do not need to be considered.

Likewise you could assume that your system consists of all three masses acted on by external force i, $(F)$.

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A string pulls both the blocks attached to it with equal amounts of force . (Given by the Tension in the string .)

The illustration is a little confusing actually . The net force acting on the second block is T2- T1.

This result is only for a massless string

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  • $\begingroup$ Right. But what about the 40 N force applied to the right? Wouldn't that add up in the equation? A free body diagram would be a great help. Thanks. $\endgroup$ – Ritwik Ojha Apr 11 '18 at 6:06
  • $\begingroup$ @Ritwik ... For that , $F-T=ma$ ... $\endgroup$ – Nehal Samee Apr 11 '18 at 6:39
  • $\begingroup$ Draw the free body diagrams for all the blocks independently , The force 40 N and T2 appears in the Free body diagram for the 1st block only . I'll give the FBD's as an edit . $\endgroup$ – feynmanfanboy Apr 11 '18 at 6:55
  • $\begingroup$ The force T2 and T1 appear in the second block 's FBD and only T1 appears in the third's . $\endgroup$ – feynmanfanboy Apr 11 '18 at 7:03

protected by Qmechanic Apr 11 '18 at 11:10

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