Question

Two forces P and 2P are acting at an angle theta. If the square of the resultant of them is equal to half the sum of the squares of their magnitude then theta is ?

Answer 1

Hint:Use the formula to determine the resultant of the two vectors. This formula gives the relation between the magnitudes of two vectors and angle between the two vectors. Then use the condition for the resultant of the given forces and determine the magnitude of angle between the two forces.

Formula used:
The resultant of two vectors is given by
\[R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \] …… (1)
Here, \[A\] and \[B\] are the magnitudes of the vectors \[\vec A\] and \[\vec B\] respectively and is the angle between the vectors \[\vec A\] and \[\vec B\].

Complete step by step answer:
We have given that there are two forces \[P\] and \[2P\] making an angle \[\theta \] with each other. Let \[R\] be the resultant of the two forces. Let us first determine the resultant of the two given forces using equation (1).
Substitute \[P\] for \[A\] and \[2P\] for \[B\] in equation (1).
\[R = \sqrt {{P^2} + {{\left( {2P} \right)}^2} + 2P\left( {2P} \right)\cos \theta } \]
\[ \Rightarrow R = \sqrt {{P^2} + 4{P^2} + 4{P^2}\cos \theta } \]
\[ \Rightarrow R = \sqrt {5{P^2} + 4{P^2}\cos \theta } \]

We have given that the square of resultant \[R\] of the two forces is equal to half the sum of the squares of their magnitude.
\[{R^2} = \dfrac{1}{2}\left[ {{P^2} + {{\left( {2P} \right)}^2}} \right]\]

Substitute \[\sqrt {5{P^2} + 4{P^2}\cos \theta } \] for \[R\] in the above equation.
\[{\left( {\sqrt {5{P^2} + 4{P^2}\cos \theta } } \right)^2} = \dfrac{1}{2}\left[ {{P^2} + {{\left( {2P} \right)}^2}} \right]\]
\[ \Rightarrow 5{P^2} + 4{P^2}\cos \theta = \dfrac{1}{2}\left[ {{P^2} + 4{P^2}} \right]\]
\[ \Rightarrow 4{P^2}\cos \theta = \dfrac{{5{P^2}}}{2} - 5{P^2}\]
\[ \Rightarrow 4{P^2}\cos \theta = \dfrac{{5{P^2} - 10{P^2}}}{2}\]
\[ \Rightarrow 4{P^2}\cos \theta = \dfrac{{ - 5{P^2}}}{2}\]
\[ \Rightarrow \cos \theta = \dfrac{{ - 5}}{8}\]
\[ \therefore \theta = {\cos ^{ - 1}}\left( {\dfrac{5}{8}} \right)\]

Hence, the value of theta is equal to \[{\cos ^{ - 1}}\left( {\dfrac{5}{8}} \right)\].

Additional information:
If the angle between the two vectors is \[90^\circ \] then the equation (1) for the resultant of two vectors becomes
\[R = \sqrt {{A^2} + {B^2}} \]
If the angle between the two vectors is \[0^\circ \] then the equation (1) for the resultant of two vectors becomes
\[R = \sqrt {{A^2} + {B^2} + 2AB} \]

Note: The students may wonder why the negative sign in the final answer for the value of theta is removed. But we know that the cosine of negative of an angle is equal to the cosine of the same positive angle. Hence, the answer with cosine inverse of positive or negative of the same angle gives the same answer.