On a pulley of mass M hangs a rope with two m | vedclass

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(C) Since the pulley is massive,the tensions on both sides of the pulley are not equal. From the free body diagram,we have:$m_{1} g - T_{1} = m_{1} a$

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(C) Since the pulley is massive,the tensions on both sides of the pulley are not equal. From the free body diagram,we have:$m_{1} g - T_{1} = m_{1} a$  ---$(i)$$T_{2} - m_{2} g = m_{2} a$  ---(ii)$(T_{1} - T_{2}) R = I \alpha = \left(\frac{M R^{2}}{2}\right) \left(\frac{a}{R}\right) = \frac{M R a}{2}$  ---(iii)From equation (iii),we get:$T_{1} - T_{2} = \frac{M a}{2}$  ---(iv)Substituting $T_{1} = m_{1}(g - a)$ and $T_{2} = m_{2}(g + a)$ into equation (iii):$m_{1}(g - a) - m_{2}(g + a) = \frac{M a}{2}$$(m_{1} - m_{2}) g = a \left(\frac{M}{2} + m_{1} + m_{2}\right)$$a = \frac{(m_{1} - m_{2}) g}{\frac{M}{2} + (m_{1} + m_{2})}$Substituting $a$ into equation (iv):$T_{1} - T_{2} = \frac{M}{2} \cdot \frac{(m_{1} - m_{2}) g}{\frac{M}{2} + (m_{1} + m_{2})} = \frac{(m_{1} - m_{2}) g}{1 + \frac{2(m_{1} + m_{2})}{M}}$As $M \rightarrow 0$,$T_{1} - T_{2} \rightarrow 0$. As $M \rightarrow \infty$,$T_{1} - T_{2} \rightarrow (m_{1} - m_{2}) g$. The expression shows that the difference increases with $M$ and approaches a constant value,which corresponds to the curve in graph $(c)$.

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$A$ $\sqrt{34}\,m$ long ladder weighing $10\,kg$ leans on a frictionless wall. Its feet rest on the floor $3\,m$ away from the wall as shown in the figure. If $F_{f}$ and $F_{w}$ are the reaction forces of the floor and the wall,then the ratio of $F_{w}/F_{f}$ will be:
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