First slide

Motion in a plane

Question

The resultant of two forces 2P and $\large \sqrt{2}P$ is $\large \sqrt{10}P$ . The angle between the forces is

Moderate

A

30⁰
B

60⁰
C

45⁰
D

90⁰

Solution

$R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta }$
$\sqrt {10} P = \sqrt {4{P^2} + 2{P^2} + 2 \times 2P \times \sqrt 2 P\cos \theta }$
$10{P^2} = 6{P^2} + 4\sqrt 2 {P^2}\cos \theta$
$4 = 4\sqrt 2 \cos \theta \Rightarrow \cos \theta = \frac{1}{{\sqrt 2 }} = \cos 45$
$\theta = {45^0}$

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