Question

A load of 981 N is suspended from a steel wire of radius 1 mm. what is the maximum angle through which the wire with the load can be deflected so that it does not break when the load passes through the equilibrium poision? Breaking stress is 7.85 xx 10^(8)Nm^(-2).

Answer 1

Answer:

sorry I didn't know what You mean

Answer 2

Answer:

θ =75.9°..

Explanation:

Let θ be the maximum angular deflection them 1/2mv2 = mg (l - lcosθ) and T - mg = mv2/l T = mg + mv2/l T/πr2 = mg + 2 mg (l - lcosθ) = 3mg - 2mg cosθ For maximum value of θ, stress (=T/πr2) must be equal to breaking stress, i.e., T/πr2 = 7.85 x 108 or, (3mg - 2mg cosθ)/(π x (10-3)2) = 7.85 x 108 or, cosθ = 0.2436 = cos 75.9° or, θ =75.9°..