First slide

Motion in a plane

Question

The resultant of two equal forces is 141.4 N when they are mutually perpendicular. When they are inclined at an angle 120º, then the resultant force will be

Moderate

Solution

$\large R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta }$
$\large P = Q,\;\theta = {90^0};\;R = \sqrt 2 P,R = 141.1N$
$\large \therefore 141.4 = 1.414P$
$\large P = 100N = Q,\theta = {120^0}$
$\large {R^1} = \sqrt {{P^2} + {P^2} + 2{P^2}Cos{\theta ^1}}$
$\large = \sqrt {{{(100)}^2} + {{(100)}^2} + 2{{(100)}^2}\left( { - \frac{1}{2}} \right)}$
$\large \left( {\cos {{120}^0} = - \frac{1}{2}} \right) = 100N$

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