A mass M is suspended by a rope from rigid support at A as shown in the figure. Another rope is tied at the end B and it is pulled horizontally with a force F . If the rope AB makes an angle θ with the vertical, then the tension in the string AB is

Question

A mass M is suspended by a rope from rigid support at A as shown in the figure. Another rope is tied at the end B and it is pulled horizontally with a force F . If the rope AB makes an angle θ with the vertical, then the tension in the string AB is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yl8onz7dg7rha3u9.jpg" /> (A) Fsin θ  (B) (F/s i n θ ) (C) Fcos θ  (D) (F/c o s θ )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-yl8onz7dg7rhjn7o.jpg" /> (T/sin (90)o)=(F/sin â¡ (π - θ ))=(M g/sin â¡ ((π )2 + θ )) Or T=(F/s i n θ )=(M g/c o s θ )

Q. A mass $M$ is suspended by a rope from rigid support at $A$ as shown in the figure. Another rope is tied at the end $B$ and it is pulled horizontally with a force $F$ . If the rope $A B$ makes an angle $θ$ with the vertical, then the tension in the string $A B$ is

$F s in θ$

$\frac{F}{s in θ}$

$F cos θ$

$\frac{F}{cos θ}$

$\frac{T}{s in ( 90 ) ^{o}} = \frac{F}{s in ( π - θ )} = \frac{M g}{s in ( \frac{π}{2} + θ )}$
Or $T = \frac{F}{s in θ} = \frac{M g}{cos θ}$

Q. A mass M is suspended by a rope from rigid support at A as shown in the figure. Another rope is tied at the end B and it is pulled horizontally with a force F . If the rope AB makes an angle θ with the vertical, then the tension in the string AB is

Fsinθ

sinθF​

Fcosθ

cosθF​