Question 11: Breaking Force of Rod
Concept: When a rod is pulled down, the upper part develops compressive stress while the lower part develops tensile stress. The stress condition for breaking remains the same if the Torque about the point is the same in the frame of the rod.
1. System Dynamics Equation:
Consider the forces acting on the rod mass \( M \) and attached masses \( m \). $$ F – 2T = Ma $$ $$ T = ma $$ Substituting \( T \): $$ F – 2(ma) = Ma $$ $$ F = (M + 2m)a \implies a = \frac{F}{M + 2m} $$
Consider the forces acting on the rod mass \( M \) and attached masses \( m \). $$ F – 2T = Ma $$ $$ T = ma $$ Substituting \( T \): $$ F – 2(ma) = Ma $$ $$ F = (M + 2m)a \implies a = \frac{F}{M + 2m} $$
2. Torque Balance (Moment Equation):
Equating the internal moment required to break the rod to the static breaking condition \( F_0 \). $$ (ma)x + \left(\frac{Ma}{2}\right) \frac{x}{2} = \frac{F_0}{2} x $$ $$ ma + \frac{Ma}{4} = \frac{F_0}{2} $$ $$ a \left( \frac{4m + M}{4} \right) = \frac{F_0}{2} $$ $$ a = \frac{2F_0}{4m + M} $$
Equating the internal moment required to break the rod to the static breaking condition \( F_0 \). $$ (ma)x + \left(\frac{Ma}{2}\right) \frac{x}{2} = \frac{F_0}{2} x $$ $$ ma + \frac{Ma}{4} = \frac{F_0}{2} $$ $$ a \left( \frac{4m + M}{4} \right) = \frac{F_0}{2} $$ $$ a = \frac{2F_0}{4m + M} $$
3. Final Calculation:
Equating the acceleration \( a \) from both steps: $$ \frac{F}{M + 2m} = \frac{2F_0}{4m + M} $$ $$ F = \frac{2 F_0 (M + 2m)}{4m + M} $$
Equating the acceleration \( a \) from both steps: $$ \frac{F}{M + 2m} = \frac{2F_0}{4m + M} $$ $$ F = \frac{2 F_0 (M + 2m)}{4m + M} $$