A body of mass 1 kg is suspended with the he | vedclass

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

Question

(B) For the body to be in equilibrium,the net force in both horizontal and vertical directions must be zero.Let $m = 1 \ kg$ and $g = 10 \ m/s^2$. The

Vedclass · Accepted Answer

$5 \sqrt{3}, 5$

Vedclass · Answer

(B) For the body to be in equilibrium,the net force in both horizontal and vertical directions must be zero.Let $m = 1 \ kg$ and $g = 10 \ m/s^2$. The weight $W = mg = 1 	imes 10 = 10 \ N$.Resolving the tensions into components:Horizontal direction: $T_1 \cos 60^{\circ} = T_2 \cos 30^{\circ}$$T_1 (1/2) = T_2 (\sqrt{3}/2) \implies T_1 = T_2 \sqrt{3}$Vertical direction: $T_1 \sin 60^{\circ} + T_2 \sin 30^{\circ} = mg$$T_1 (\sqrt{3}/2) + T_2 (1/2) = 10$Substituting $T_1 = T_2 \sqrt{3}$ into the vertical equation:$(T_2 \sqrt{3}) (\sqrt{3}/2) + T_2 (1/2) = 10$$T_2 (3/2) + T_2 (1/2) = 10$$2 T_2 = 10 \implies T_2 = 5 \ N$Now,$T_1 = T_2 \sqrt{3} = 5 \sqrt{3} \ N$.Thus,the tensions $T_1$ and $T_2$ are $5 \sqrt{3} \ N$ and $5 \ N$ respectively.

$A$ body of mass $1 \ kg$ is suspended with the help of two strings making angles as shown in the figure. The magnitudes of tensions $T_1$ and $T_2$,respectively,are (in $N$):

Similar Questions

If more than one force acts on a particle,state the condition for the equilibrium of the particle.

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

$A$ block of $\sqrt{3} \, kg$ is attached to a string whose other end is attached to the wall. An unknown force $F$ is applied so that the string makes an angle of $30^{\circ}$ with the wall. The tension $T$ is $........... \, N$ : (Given $g = 10 \, ms^{-2}$)

In the given arrangement,the normal force applied by the block on the ground is

$A$ man of mass $50 \,kg$ carries a bag of weight $40 \,N$ on his shoulder. The force with which the floor pushes up his feet will be ......... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module