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There are two bodies viz. A and B which are placed on a horizontal surface. The free end P is being pulled by a constant force of 1 N. Find the acceleration of free end P?

Figure

My approach

Sign convention -––> Left is considered to be negative and right is considered to be positive.

String is considered to be massless.

So at point P,

$F - T = ma = 0$

=> $F = T = 1 N$

According to my perception,

The tension is same throughout the string.

So the block A is in contact at two points.

So overall Force that is pulling the block A is 3 N. (Due to Tension force acting on the pulleys)

Similarly, The force acting on Block B is -4 N.

So both the blocks must have an acceleration of 1m/s^2. (A towards right and B towards left)

So now using,

$ a_{pulley} = \frac{a_{string1} + a_{string2}}{2}$

figure 2

Using this,

$a_{string} at point p = 7\frac{m}{s^2}$

Is this the most efficient way to solve this problem?

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closed as off-topic by Michael Seifert, Bob D, Aaron Stevens, StephenG, John Rennie Jul 9 at 4:44

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    $\begingroup$ Show some effort ie what you have worked out so far, then people will help. $\endgroup$ – user207455 Jul 8 at 12:13
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    $\begingroup$ I shared my effort. $\endgroup$ – Kaushik Jul 8 at 15:38
  • $\begingroup$ Do Newton's 2nd law on both boxes. When you know those accelerations, some geometry considerations might tell the string acceleration at different points on the string. $\endgroup$ – Steeven Jul 8 at 17:07
  • $\begingroup$ @Steeven I arrived at a solution. But is this the most efficient way to solve this problem? $\endgroup$ – Kaushik Jul 8 at 19:41