Question

A $60kg$ man standing on a massless weighing machine, placed on a $30kg$ plank keeps himself stable by pulling the ropes $B,C$ as shown in the figure. Assume the ropes and pulleys are light and neglect friction at all the surfaces. The plank remains horizontal. To keep himself stable, the man has to apply a total force of ...... $N$ on ropes $B$ and $C$, the reading of the weighing machine then is ....... kg.

• A. $900,90$
• B. $400,40$
• C. $300,40$
• D. $400,50$

Answer 1

The correct option is A $900,90$

Let the tension force be $T$ in the rope$.$ The downward forces acting on the system of the man and the box are the weight of the man and the weight of the box$.$ The upward forces are $T$ and $T$ on each side of the rope$.$

As the system is in a static equilibrium$:$

$T+T=(60+30)g$

$=>T=45kgf$ or $45gN$ $=450N$ if $g=10m/s^2$

The forces acting on the man are Tension $T,$ reaction from the weighing machine$,$ and weight $60g.$

so$,$ Normal reaction of the weighing machine $=60-T=15kgf.$

$\left(i\right)$ The weighing machine shows its normal reaction force as the weight of the person standing on it$.$ So weight displayed $=15kg.$

$(ii$) Suppose the man pulls down the rope with a force $F$ and the tension becomes $T.$

The forces acting downwards on the rope at the two ends are$:F,$ weights of man and the box$.$ The upward forces are Tension $T$ on each side of rope$.$

The system being in equilibrium$:2T=F+60kgf+30kgf=F+90kgf$

$=>T=45kgf+F/2$

$=>$ Normal reaction force from the weighing machine on the man

$N=60+F-T=15kgf+F/2$

If $N$ is to become $60kgf,$ then $F$ has to be $90kgf$ or $900N$

Hence,

option $\left(A\right)$ is correct answer.