A uniform rod of length L and mass M is place | vedclass

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(B) Let the rod extend from $x = 0$ to $x = L$. The force $F$ is applied at $x = L$. The friction force on an element $dx$ at position $x$ is $df = \m

Vedclass · Accepted Answer

$F/4$

Vedclass · Answer

(B) Let the rod extend from $x = 0$ to $x = L$. The force $F$ is applied at $x = L$. The friction force on an element $dx$ at position $x$ is $df = \mu(x) \cdot dm \cdot g = (Kx) \cdot (M/L) dx \cdot g$.The total friction force $F$ required to keep the rod in equilibrium is the integral of $df$ from $0$ to $L$:$F = \int_{0}^{L} \frac{MgK}{L} x dx = \frac{MgK}{L} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{MgKL}{2}$.Thus,$K = \frac{2F}{MgL}$.Now,consider the tension $T$ at the midpoint $x = L/2$. The tension $T$ must balance the total friction force acting on the segment from $x = 0$ to $x = L/2$:$T = \int_{0}^{L/2} \frac{MgK}{L} x dx = \frac{MgK}{L} \left[ \frac{x^2}{2} \right]_{0}^{L/2} = \frac{MgK}{L} \cdot \frac{L^2}{8} = \frac{MgKL}{8}$.Substituting $K = \frac{2F}{MgL}$ into the expression for $T$:$T = \frac{MgL}{8} \cdot \left( \frac{2F}{MgL} \right) = \frac{2F}{8} = \frac{F}{4}$.

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