Question:
Forces 5 N, 12 N and 13 N are in equilibrium. If $ sin\text{ }23{}^\circ =\frac{5}{13}, $ the angle between 5 N force and 13 N force is
Forces 5 N, 12 N and 13 N are in equilibrium. If $ sin\text{ }23{}^\circ =\frac{5}{13}, $ the angle between 5 N force and 13 N force is
Updated On: Aug 15, 2022
- $ 23{}^\circ $
- $ 67{}^\circ $
- $ 90{}^\circ $
- $ 113{}^\circ $
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The Correct Option is B
Solution and Explanation
As $ {{5}^{2}}+{{12}^{2}}={{13}^{2}}, $ so these forces will be acting along the sides of a right angled triangle. Angle between 5 N and 13 N
$=180-(90+23) $
$=180-113={{67}^{o}} $
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration