A load of $$981$$ N is suspended from a steel wire of radius $$1mm$$. The maximum angle through which the wire with a load can be deflected so that it does not break when the load passes through the position of equilibrium is computed as $$cos$$−$$1$$[α(0.081)]. Find α . (Breaking$$ stress is $$7.85$$×$$108N/m2$$ $$)

Question

A load of $$981$$ N is suspended from a steel wire of radius $$1mm$$. The maximum angle through which the wire with a load can be deflected so that it does not break when the load passes through the position of equilibrium is computed as $$cos$$−$$1$$[α(0.081)].  Find α . (Breaking$$ stress is $$7.85$$×$$108N/m2$$ $$)

Infinity Learn · Accepted Answer

$$TmaxA=mgA(3$$−$$2cos$$θ$$)=7.85$$×$$108$$       $$981$$×$$722$$×$$10$$−$$6$$[$$3$$−$$2$$  $$cos$$θ]$$=7.85$$×$$108$$     $$312.13$$×$$106$$[$$3$$−$$2$$  $$cos$$θ]$$=7.85$$×$$108$$     ∴θ$$=3$$

Elastic behaviour of solids

Remember concepts with our Masterclasses.

$\frac{T_{\max}}{A} = \frac{m g}{A} (3 - 2 \cos θ) = 7.85 \times 10^{8}$
$\frac{981 \times 7}{22 \times 10^{- 6}} [3 - 2 \cos θ] = 7.85 \times 10^{8}$

$312.13 \times 10^{6} [3 - 2 \cos θ] = 7.85 \times 10^{8}$

$∴ θ = 3$

Ready to Test Your Skills?

free study material

Elastic behaviour of solids

Remember concepts with our Masterclasses.

A load of 981 N is suspended from a steel wire of radius 1mm. The maximum angle through which the wire with a load can be deflected so that it does not break when the load passes through the position of equilibrium is computed as cos−1[α(0.081)]. Find α . (Breaking stress is 7.85×108N/m2 )

TmaxA=mgA(3−2cosθ)=7.85×108 981×722×10−6[3−2 cosθ]=7.85×108 312.13×106[3−2 cosθ]=7.85×108 ∴θ=3

Ready to Test Your Skills?

free study material

$\frac{T_{\max}}{A} = \frac{m g}{A} (3 - 2 \cos θ) = 7.85 \times 10^{8}$
$\frac{981 \times 7}{22 \times 10^{- 6}} [3 - 2 \cos θ] = 7.85 \times 10^{8}$

$312.13 \times 10^{6} [3 - 2 \cos θ] = 7.85 \times 10^{8}$

$∴ θ = 3$