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

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.