Question

The wires A and B shown in the figure are made of the same material, and have radii r_A and r_B respectively. The block between them has a mass m. When the force F is mg/3, one of the wires breaks.

• A. A will break before B if $r_A=r_B$
• B. A will break before B if $r_A=r_B$
• C. Either A or B may break if $r_A<2r_B$
• D. The lengths of A and B must be known to predict which wire will break

Answer 1

Answer: (A), (B), (C)

Either A or B may break if $r_A<2r_B$

Tension in wire A = $Mg+\frac{Mg}{3}$

Tension in wire B = $\frac{Mg}{3}$
Stress in A, $S_A=\frac{T}{A}=\frac{4}{3}\frac{Mg}{\pi r_R^2}$
Stress in B, $S_B=\frac{Mg}{\pi r_R^2}$
We know that one of the wires breaks when = $F=\frac{mg}{3}$
Which means if $r_A=r_B$ then A should break as stress in A is greater.
We can also see that $S_A$ will be always greater unless $r_A>2r_B$ which means that A will break
if $r_A<2r_B$
Similarly, when $r_A=2r_B$ stress will be the same on both wires, therefore either can break.