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).

A load of 981 N is suspended from a steel wire of radius 1 mm. what is

1.	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» SOLUTION :Let `theta` be the maximum angular deflection of load and `upsilon` be the velocity ACQUIRED by load while passing through the EQUILIBRIUM POSITION A. PE of load at `B = KE ` of load at A `mg(L-l cos theta) = 1.2 m upsilon^(2)` …(i) If T is the tension in the string at the lowest position A, then `T - mg = (m upsilon^(2))/(l)` or `T = mg +(m upsilon^(2))/(l) = mg + 2 m (1- cos theta)` `= 3mg - 2 mg cos theta` For maximum value of `theta` stress = `(T//pi r^(2))` must be equal to breaking stress i.e., `(T)/(pi r^(2)) = 7.85 xx 10^(8)` or`(3 mg - 2mg cos theta)/(pi r^(2)) = 7.85 xx 10^(8)` or `(3 xx 981 - 2 xx 981 cos theta)/((22//7)xx(10^(-3))^(2)) = 7.85 xx 10^(8)` or `2943-1962 cos theta = (22//7) xx 10^(-6) xx 7.85 xx 10^(8)` `=2466.5` or `cos theta = (2943-2466.5)/(1962) = (476.5)/(1962)` `= 0.2429 = cos 75^(@) 57'` `theta = 75^(@) 57'`.