As depicted in the figure, a man of mass 60 kg stands on a light...

Let the mass of the man be $M = 60 \text{ kg} and the mass of the box be$m = 30 \text{ kg}$. Let$T$be the tension in the rope and$N$be the normal reaction between the man and the weighing machine.<br><br>When the box is stationary, the acceleration of the system is zero. Considering the man and the box as a single system, the upward forces are the tensions from the two segments of the rope ($2T$), and the downward force is the total weight.<br><br>$2T = (M + m)g$<br><br>$2T = (60 + 30)g = 90g \Rightarrow T = 45g$<br><br>Considering the forces acting on the man alone, the upward forces are the normal reaction$N$and the tension$T$, while the downward force is his weight$Mg$.<br><br>$N + T = Mg$<br><br>$N + 45g = 60g \Rightarrow N = 15g$<br><br>The weight indicated by the weighing machine is$\dfrac{N}{g} = 15 \text{ kg}$.<br><br>When the machine displays the man's correct weight, the normal reaction is equal to the true weight of the man, so$N = 60g$. Let the system accelerate upwards with an acceleration$a$.<br><br>Applying Newton's second law for the man:<br><br>$N + T - Mg = Ma$<br><br>$60g + T - 60g = 60a \Rightarrow T = 60a \Rightarrow a = \dfrac{T}{60}$<br><br>Applying Newton's second law for the entire system:<br><br>$2T - (M + m)g = (M + m)a$<br><br>$2T - 90g = 90a$<br><br>Substituting$a = \dfrac{T}{60}$:<br><br>$2T - 90g = 90 \left(\dfrac{T}{60}\right)

$2T - 90g = 1.5T$

$0.5T = 90g \Rightarrow T = 180g$

Taking $g = 10 \text{ m/s}^2$, the force the man must exert on the rope is:

$T = 180 \times 10 = 1800 \text{ N}$.