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# $P,Q$ $and{\text{ }}R$ are three coplanar forces acting at a point and they are in equilibrium. Given$P = 1.9318Kg{\text{ }}wt$ , $\sin {\theta _1} = 0.9659$ , Find the value of $R\left( {in{\text{ }}Kg{\text{ }}wt} \right)$.

Last updated date: 05th Mar 2024
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Hint: For solving this problem we need to have a clear idea of Lami’s Theorem.
According to this theorem when 3 forces are acting at a point and they are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.
Lami’s theorem is useful for determining the magnitude of unknown forces for a given system.
With this theorem, the property of 3 forces in equilibrium is considered to be a very neat method of solving any kind of three-force problem. If we know the angles between pairs of forces the best method to find these out is Lami’s theorem.
The 3 forces should be inward or outward and opposite.

Formula used:
Simply follow Lami’s Theorem formula in this case
$\dfrac{A}{{\operatorname{Sin} \alpha }} = \dfrac{B}{{\operatorname{Sin} \beta }} = \dfrac{C}{{\operatorname{Sin} \gamma }}$

Complete step-by-step solution:

According to the statement of this theorem when 3 forces are acting at a point and they are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.
$\dfrac{A}{{\operatorname{Sin} \alpha }} = \dfrac{B}{{\operatorname{Sin} \beta }} = \dfrac{C}{{\operatorname{Sin} \gamma }}$
Using the above formula for Lami’s theorem
$\dfrac{P}{{\sin {\theta _1}}} = \dfrac{Q}{{\sin {\theta _2}}} = \dfrac{R}{{\sin 150}}$
$\therefore \dfrac{P}{{\sin {\theta _1}}} = \dfrac{R}{{\operatorname{Sin} 150}}$
Put values P=1.9318 Kg wt, $\sin {\theta _1} = 0.9659$ in the above equation we get,
$\Rightarrow \dfrac{{1.9318}}{{0.9659}} = \dfrac{R}{{0.5}}$
$\therefore \dfrac{{1.9318 \times 0.5}}{{0.9659}} = R$
So, finally the value of$R = 1Kg{\text{ }}wt$ .