A mass of 10 , kg is suspended vertically by | vedclass

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

Question

(D) Let the tension in the upper half of the rope be $T_1$ and the tension in the lower half be $T_2$. The lower half of the rope supports the mass $m

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the tension in the upper half of the rope be $T_1$ and the tension in the lower half be $T_2$. The lower half of the rope supports the mass $m = 10 \, kg$,so $T_2 = mg = 10 	imes 10 = 100 \, N$.At the middle point where the horizontal force $F = 30 \, N$ is applied,we consider the equilibrium of forces in the horizontal and vertical directions.For horizontal equilibrium: $T_1 \sin 	heta = F = 30 \, N$.For vertical equilibrium: $T_1 \cos 	heta = T_2 = 100 \, N$.Dividing the two equations: $\frac{T_1 \sin 	heta}{T_1 \cos 	heta} = \frac{30}{100} \Rightarrow 	an 	heta = 0.3$.Given $	heta = 	an^{-1}(x 	imes 10^{-1})$,we have $	an 	heta = x 	imes 10^{-1} = \frac{x}{10}$.Equating the two expressions for $	an 	heta$: $\frac{x}{10} = 0.3 \Rightarrow x = 3$.

Similar Questions

$A$ body of weight $2 \, kg-wt$ is suspended as shown in the figure. The tension $T_1$ in the horizontal string (in $kg-wt$) is

$A$ $50 \, kg$ homogeneous smooth sphere rests on a $30^{\circ}$ incline $A$ and bears against a smooth vertical wall $B$. Calculate the contact force at $A$. (Take $g = 10 \, m/s^2$)

$A$ box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $F_1$ is $10 \ N$,what is the magnitude of $F_3$ (in $N$)?

When three forces of $50 \, N$,$30 \, N$,and $15 \, N$ act on a body,then the body is

$A$ man is standing on a balance and his weight is measured. If he takes a step to the left side,then his weight:

Vedclass Test Series

Exam Paper Generator

Online Exam Module