The breaking stress of a wire going over a sm | vedclass

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(B) For the system shown,the equation of motion for the $2\, kg$ mass is: $2g - T = 2a \Rightarrow 20 - T = 2a$  $...(1)$For the $1\, kg$ mass,the equ

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$0.46 	imes 10^{-4}\, m$

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(B) For the system shown,the equation of motion for the $2\, kg$ mass is: $2g - T = 2a \Rightarrow 20 - T = 2a$  $...(1)$For the $1\, kg$ mass,the equation of motion is: $T - 1g = 1a \Rightarrow T - 10 = a$  $...(2)$Adding equations $(1)$ and $(2)$: $(20 - T) + (T - 10) = 2a + a \Rightarrow 10 = 3a \Rightarrow a = \frac{10}{3}\, m/s^2$.Substituting $a$ into equation $(2)$: $T = 10 + a = 10 + \frac{10}{3} = \frac{40}{3}\, N$.The breaking stress is given by $	ext{Stress} = \frac{T}{A} = 2 	imes 10^9\, N/m^2$.Therefore,$A = \frac{T}{2 	imes 10^9} = \frac{40/3}{2 	imes 10^9} = \frac{20}{3} 	imes 10^{-9}\, m^2$.Since $A = \pi r^2$,we have $\pi r^2 = \frac{20}{3} 	imes 10^{-9}$.$r^2 = \frac{20}{3 	imes 3.14} 	imes 10^{-9} \approx 2.123 	imes 10^{-10} = 21.23 	imes 10^{-11}$.$r = \sqrt{21.23} 	imes 10^{-5.5} \approx 4.6 	imes 10^{-6} 	imes 10 = 0.46 	imes 10^{-4}\, m$.

The breaking stress of a wire going over a smooth pulley is $2 \times 10^9\, N/m^2$. What would be the minimum radius of the wire used if it is not to break? (Assume $g = 10\, m/s^2$)

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