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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $T$:$T = \frac{MgL}{8} \cdot \left( \frac{2F}{MgL} \right) = \frac{2F}{8} = \frac{F}{4}$.

Similar Questions

Calculate the stopping distance of a vehicle of mass $M$ moving with velocity $v$ on a straight road. (Where $\mu$ is the coefficient of friction between the tires and the road.)

$A$ body of $10\, kg$ is acted upon by a force of $129.4\, N$. If $g = 9.8\, m/s^2$ and the acceleration of the block is $10\, m/s^2$,what is the coefficient of kinetic friction?

$A$ body of mass $2\, kg$ is being dragged with uniform velocity of $2\, m/s$ on a rough horizontal plane. The coefficient of friction between the body and the surface is $0.20$. The amount of heat generated in $5\, s$ is ........ $cal$ $(J = 4.2\, J/cal$ and $g = 9.8\, m/s^2)$

$A$ body is moving along a horizontal surface with a velocity of $4 \,m/s$. If the coefficient of kinetic friction is $0.2$, the distance travelled by the body before coming to rest is $(g = 10 \,m/s^2)$. (in $\,m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module