A uniform rope of mass M and length L is pulled along a frictionless inclined plane of angle θ by applying a force F ( Mg ) parallel to incline. Tension in the chain at a distance x from the end at which force is applied is

Question

A uniform rope of mass M and length L is pulled along a frictionless inclined plane of angle θ by applying a force F (> Mg ) parallel to incline. Tension in the chain at a distance x from the end at which force is applied is (A) F ( x / L )+( x / L ) Mg sin θ (B) F (1-( x / L )) (C) F (1-( x / L ))+( x / L ) Mg sin θ (D) None of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/84e5be04e44fe8b42560cef87a67aa69-.png" /> F-M g sin θ=M a dots(1) F-((M/L) x) g sin θ-T=((M/L) x) a F-((M/L) x g sin θ)-T=(M/L) x((F-M g sin θ/M)) T=F(1-(x/L))

Q. A uniform rope of mass $M$ and length $L$ is pulled along a frictionless inclined plane of angle $θ$ by applying a force $F (> M g)$ parallel to incline. Tension in the chain at a distance $x$ from the end at which force is applied is

$F \frac{x}{L} + \frac{x}{L} M g sin θ$

$F (1 - \frac{x}{L})$

$F (1 - \frac{x}{L}) + \frac{x}{L} M g sin θ$

None of these

$F - M g sin θ = M a \dots$ (1)
$F - (\frac{M}{L} x) g sin θ - T = (\frac{M}{L} x)$ a
$F - (\frac{M}{L} xg sin θ) - T = \frac{M}{L} x (\frac{F - M g s i n θ}{M})$
$T = F (1 - \frac{x}{L})$

Q. A uniform rope of mass M and length L is pulled along a frictionless inclined plane of angle θ by applying a force F(>Mg) parallel to incline. Tension in the chain at a distance x from the end at which force is applied is

FLx​+Lx​Mgsinθ

F(1−Lx​)

F(1−Lx​)+Lx​Mgsinθ