Question

Figure shows a man of mass $60kg$ standign on a light weighing machine dept in a box of mass $30kg$. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine

Answer 1

(i) Given, Mass of man $=60kg$ [Ref. image 1]

Let $R^{\prime }=$ apparent weight of man in this case.

Now, $R^{\prime }+T-60g=0$ [From FBD of man]

$\Rightarrow T=60g-R^{\prime }$ ...(i)

$T-R^{\prime }-30g=0$ ...(ii) [From FBD of box]

$\Rightarrow 60g-R^{\prime }-R^{\prime }-30g=0$ [From (i)]

$\Rightarrow R^{\prime }=15g$ The weight shown by the machine is $15kg$.

(ii) To get his correct weight suppose the applied force is $T^{\prime }$ and so, accelerates upward with 'a'. In this case, given that correct weight $=R=60g$, where $g = acc^n due to gravity [Ref. image 2]

From the FBD of the man

$T^1+R-60g-60a=0$

$\Rightarrow T^1-60a=0$ $[\therefore R=60g]$

$\Rightarrow T^1=60a$ .....(i)

From the FBD of the box

$T^1-R-30g-30a=0$

$\Rightarrow T^1-60g-30g-30a=0$

$\Rightarrow T^1-30a=90g=900$

$\Rightarrow T^1=30a-900$ ...(ii)

From eqn. (i) and eqn.(ii) we get $T^1=2T^1-1800$

$T^1=1800N$.

$\therefore$ So, he should exact $1800N$ force on the rope to get correct reading.