Question:
At what angle must the two forces (x + y) and (x - y) act so that the resultant may be $\sqrt{x^2 + y^2}$ ?
At what angle must the two forces (x + y) and (x - y) act so that the resultant may be $\sqrt{x^2 + y^2}$ ?
Updated On: Jun 20, 2022
- $\cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]$
- $\cos^{-1} \Bigg[ - \frac{2(x^2 - y^2)}{x^2 + y^2}\Bigg]$
- $\cos^{-1} \Bigg[ - \frac{(x^2 + y^2)}{(x^2 - y^2)}\Bigg]$
- $cos^{-1} \Bigg[ - \frac{(x^2 - y^2)}{(x^2 + y^2)}\Bigg]$
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The Correct Option is A
Solution and Explanation
$R^2 = A^2 + B^2 + 2AB \, cos \, \theta$
Substituting, $A = (x + y), B = (x - y) \, and \, R = \sqrt{x^2 + y^2},$
we get, $\theta = cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]$
Substituting, $A = (x + y), B = (x - y) \, and \, R = \sqrt{x^2 + y^2},$
we get, $\theta = cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]$
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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