Two block of masses m and M are connected by means of a metal wire passing over a frictionless fixed pulley. The area of $$cross-section$$ of the wire is $$6.5$$×$$10$$−$$9m2$$ and its breaking stress is $$2$$×$$109Nm$$−$$2$$. If $$m=1kgif$$ the maximum value of M is K x $$1.86$$ kg for which the wire will not break. Find the value of K.$$g=10m/s2$$

Question

Two block of masses m and M are connected by means of a metal wire passing over a frictionless fixed pulley. The area of $$cross-section$$ of the wire is $$6.5$$×$$10$$−$$9m2$$ and its breaking stress is $$2$$×$$109Nm$$−$$2$$. If $$m=1kgif$$ the maximum value of M is K x $$1.86$$ kg for which the wire will not break. Find the value of K.$$g=10m/s2$$

Infinity Learn · Accepted Answer

Mg−$$T=Ma$$           (I)  $$T-mg$$ $$=$$ ma       (II) $$T=2Mmg(M+m)$$ ⇒$$TA=2MmgA(M+m)=2mgA1+mM=2$$×$$109$$ ⇒$$M=1.86Kg$$

Elastic behaviour of solids

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$M g - T = M a (I) T-mg = ma (II) T = \frac{2 M m g}{(M + m)} \Rightarrow \frac{T}{A} = \frac{2 M m g}{A (M + m)} = \frac{2 m g}{A (1 + \frac{m}{M})} = 2 \times 10^{9} \Rightarrow M = 1 .86 K g$

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Elastic behaviour of solids

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Mg−T=Ma (I) T-mg = ma (II) T=2Mmg(M+m) ⇒TA=2MmgA(M+m)=2mgA1+mM=2×109 ⇒M=1.86Kg

Ready to Test Your Skills?

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$M g - T = M a (I) T-mg = ma (II) T = \frac{2 M m g}{(M + m)} \Rightarrow \frac{T}{A} = \frac{2 M m g}{A (M + m)} = \frac{2 m g}{A (1 + \frac{m}{M})} = 2 \times 10^{9} \Rightarrow M = 1 .86 K g$