A mass of 10 ; kg is suspended by a rope of l | vedclass

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

Question

(A) Let $T$ be the tension in the upper half of the rope. The lower half of the rope is vertical and supports the mass of $10 \; kg$,so the tension in

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let $T$ be the tension in the upper half of the rope. The lower half of the rope is vertical and supports the mass of $10 \; kg$,so the tension in the lower half is $T_{lower} = mg = 10 	imes 10 = 100 \; N$.At the mid-point where force $F$ is applied,we consider the equilibrium of forces.The horizontal component of the tension $T$ in the upper rope must balance the applied force $F$: $T \sin 45^{\circ} = F$.The vertical component of the tension $T$ in the upper rope must balance the downward tension from the lower rope: $T \cos 45^{\circ} = T_{lower} = 100 \; N$.Dividing the two equations: $\frac{T \sin 45^{\circ}}{T \cos 45^{\circ}} = \frac{F}{100}$.$	an 45^{\circ} = \frac{F}{100}$.Since $	an 45^{\circ} = 1$,we get $1 = \frac{F}{100}$,which implies $F = 100 \; N$.

Similar Questions

Two weights of mass $m_{1}$ and $m_{2} (> m_{1})$ are joined by an inextensible string of negligible mass passing over a fixed frictionless pulley. The magnitude of the acceleration of the loads is

$A$ painter is applying force himself to raise him and the box with an acceleration of $5\, m/s^2$ by a massless rope and pulley arrangement as shown in the figure. The mass of the painter is $100\, kg$ and that of the box is $50\, kg$. If $g = 10\, m/s^2$,then:

$A$ particle of small mass $m$ is joined to a very heavy body of mass $M$ by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is

Vedclass Test Series

Exam Paper Generator

Online Exam Module