Question

A man of mass $m$ stands on a platform of equal mass $m$ and pulls himself by two ropes passing over pulleys as shown in the figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be:

• A. $\frac{g}{2}$
• B. $\frac{g}{4}$
• C. $g$
• D. $zero$

Answer 1

The correct option is D $zero$

$\text{Step 1: Drawing F.B.D and Assuming System [Refer Fig. 1 and 2]}$

Assuming Man plus Platform as system of mass 2m. Drawing its FBD in Fig. 2.

We need not mention the internal force Normal Reaction $N$ between man and platform, as they will cancel out. Therefore, only External forces are mentioned.

$\text{Step 2: Finding Tension in string by Newton's 3rd Law}$
As the Man pulls each rope, the tension in the string equals the pull of the man. As tension and his pull will be action reaction pair by Newton's third law.

$\Rightarrow T=\frac{mg}{2}$ ( Since, Pull of man equals half of his weight)

Also, As the string is lignt and pulleys are massless, therefore the Tension on same string is same at every point [Refer Fig. 1]

$\text{Step 3: Applying Newton's Second Law}$

On (Man + platform) [Refer Fig. 2]

In y direction, upwards positive:

$\Sigma F_y=m^{\prime }a,$ Where $m^{\prime }=2m$

$4T-2mg=2ma$

$4\left(\frac{mg}{2}\right)-2mg=2ma$ $\Rightarrow a=0$

So acceleration of system will be zero.

Option D is correct.