P, Q and R are three coplanar forces acting a | vedclass

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

Question

(C) According to Lami's Theorem,for three forces in equilibrium:
$\frac{P}{\sin 	heta_R} = \frac{Q}{\sin 	heta_P} = \frac{R}{\sin 	heta_Q}$
F

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) According to Lami's Theorem,for three forces in equilibrium:
$\frac{P}{\sin 	heta_R} = \frac{Q}{\sin 	heta_P} = \frac{R}{\sin 	heta_Q}$
From the figure,the angle opposite to $P$ is $	heta_1$,the angle opposite to $Q$ is $	heta_2$,and the angle opposite to $R$ is $150^\circ$.
Thus,$\frac{P}{\sin 	heta_1} = \frac{R}{\sin 150^\circ}$.
Given $P = 1.9318 \, kg \, wt$ and $\sin 	heta_1 = 0.9659$.
Substituting the values:
$\frac{1.9318}{0.9659} = \frac{R}{\sin 150^\circ}$
Since $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = 0.5$.
$2 = \frac{R}{0.5}$
$R = 2 	imes 0.5 = 1 \, kg \, wt$.

$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P = 1.9318 \, kg \, wt$,$\sin {\theta _1} = 0.9659$,the value of $R$ is (in $kg \, wt$):

Similar Questions

There are four forces acting at a point $P$ produced by strings as shown in the figure. The system is at rest. Find the forces $F_1$ and $F_2$.

As shown in the figure,the tension in the horizontal cord is $30 \, N$. The weight $W$ and the tension in the string $OA$ in Newton are:

When an object is in equilibrium state,then

For the free body diagram shown in the figure,four forces are applied in the $x$ and $y$ directions. What additional force must be applied and at what angle with the positive $x$-axis so that the net acceleration of the body is zero?

$A$ body is moving under the action of two forces $\vec{F}_1 = 2\hat{i} - 5\hat{j}$ and $\vec{F}_2 = 3\hat{i} - 4\hat{j}$. Its velocity will become uniform under an additional third force $\vec{F}_3$ given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module