Question

Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle 𝛉 with each other. If the force Q is doubled, then their resultant also gets doubled. then, the angle is
A. 30˚
B. 60˚
C. 90˚
D. 120˚

Answer 1

Hint: we will use the formula, to find the magnitude of resultant of two vectors.

Complete step by step solution:
In the question we are given two forces P and Q of magnitude;
\[P = 2F\]and \[Q = 3F\]and the angle between the vectors is 𝛉, then it is said that if force Q is doubled then their resultant also get doubled so, we required to find the value of θ
First let us draw the diagram of these vectors

 let the resultant of vector P and Q is vector R
SO, \[\overrightarrow P + \overrightarrow Q = \overrightarrow R \] and |\[\overrightarrow P + \overrightarrow Q = |R|\]=\[\sqrt {{P^2} + {Q^2} + 2PQCOS\theta } \]
Here, \[P = 2F\]and \[Q = 3F\] so put the value in above equation
\[R = \sqrt {{{(2F)}^2} + {{(3F)}^2} + 2(2F)(3F)COS\theta } \] =\[\sqrt {13{{(F)}^2} + 12{{(F)}^2}COS\theta } \]
 now the vectors Q gets doubled, so, Q= 6F and the resultant vector also get doubled, so ,
 \[{R_1} = 2\sqrt {13{{(F)}^2} + 12{{(F)}^2}COS\theta } \] …………..(1)
And the magnitude of new resultant is given as \[{R_1} = \sqrt {{{(2F)}^2} + {{(6F)}^2} + 2(2F)(6F)COS\theta } \]
                                                                                         =\[\sqrt {40{{(F)}^2} + 24{{(F)}^2}COS\theta } \]
From equation 1 \[\sqrt {40{{(F)}^2} + 24{{(F)}^2}COS\theta } \] =\[2\sqrt {13{{(F)}^2} + 12{{(F)}^2}COS\theta } \]
Now squaring both side, we get
\[40{(F)^2} + 24{(F)^2}COS\theta \]=\[4\,(13{(F)^2} + 12{(F)^2}COS\theta )\]
On simplifying we get

\[40{F^2} + 24{F^2}COS\theta {\text{ }} = 52{F^2} + 48{F^2}COS\theta \]
\[ - 12{F^2} = {\text{ }}24{F^2}COS\theta \]
∴ \[cos{\text{ }}\theta {\text{ }} = \; - \dfrac{1}{2}\]

Hence, \[\theta = 120^\circ \]

Note: If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.