Question

a string tied on roof can bear maxmium tension of 50 kg wt .the minimum acceleration that can be acquired by man of 98 kg to descend will be g=9.8 m/s square

Answer 1

Answer on Question#36818 – Physics - Mechanics | Kinematics | Dynamics

A string tied on roof can bear maximum tension of $50\,\mathrm{kg}$ wt. the minimum acceleration that can be acquired by man of $98\,\mathrm{kg}$ to descend will be $g = 9.8\,\mathrm{m/s}$ square

Solution:

$$\begin{array}{l} m = 50\,\mathrm{kg} - \text{maximum tension weight of the string}; \\ M = 98\,\mathrm{kg} - \text{weight of the man}; \\ T_{\max} = mg - \text{maximum tension force of the string}; \end{array}$$

Newton's second law for the string (along Y-axis):

$$y: \mathrm{Mg} - T_{\max} = \mathrm{Ma}_{\min}$$ $$a_{\min} = \frac{\mathrm{Mg} - T_{\max}}{M} = \frac{Mg - mg}{M} = \frac{g(M - m)}{M} = \frac{9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}(98\,\mathrm{kg} - 50\,\mathrm{kg})}{98\,\mathrm{kg}} = 4.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

Answer: minimum acceleration is equal to $4.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$.