
Square of the resultant of two forces of equal magnitude is equal to three times the product of their magnitude. The angle between them is
(A) ${{0}^{\circ }}$
(B) ${{45}^{\circ }}$
(C) ${{60}^{\circ }}$
(D) ${{90}^{\circ }}$
Answer
564k+ views
Hint: From parallelogram law, we are able to find the formula for the resultant of two forces. Since we’re given that the magnitude of the forces are equal, the calculation gets easier and we can generate a definite value as output.
Formulas used:
\[R=\sqrt{A{}^\text{2}+B{}^\text{2}+2AB\cos \varphi }\],
where A and B are two forces, R is their resultant and \[\varphi \] is the angle between them.
Complete step by step answer:
Let the magnitude of force be = \[F\]
Given: \[R{}^\text{2}\text{ }=\text{ }3F{}^\text{2}\] where R is the resultant of 2 forces and \[F{}^\text{2}\] is the product of both the forces.
We have to find the angle between two forces.
Consider the angle between the equivalent forces to be \[\varphi \]
The magnitude of the resultant of the two forces is given by:
\[R=\sqrt{A{}^\text{2}+B{}^\text{2}+2AB\cos \varphi }\]
Here, \[A=B=F\]
therefore \[R=\sqrt{F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi }\]
Squaring both sides -
\[R{}^\text{2}=F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi \]
Now,
as \[R{}^\text{2}\text{ }=\text{ }3F{}^\text{2}\]
\[3F{}^\text{2}=F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi \]
or \[1=2\cos \varphi \]
Thus, \[\cos \varphi =0.5\]
\[\cos \varphi =\dfrac{1}{2}\]
\[\therefore\varphi ={{\cos }^{-1}}\dfrac{1}{2}={{60}^{\circ }}\]
Hence, the answer is option C.
Note:Even triangle law can be applied here, where the forces F and F acting at a point are shown based on magnitude and direction, by the two sides of a triangle taken in an order, and the obtained resultant is shown by the third side of the triangle which is taken in the opposite order. We have used parallelogram law where the two vectors acting at a point are shown based on magnitude and direction, by the two adjacent sides of a parallelogram generated from a particular point and the obtained resultant is represented by the diagonal of the parallelogram that is drawn from the same point.
Formulas used:
\[R=\sqrt{A{}^\text{2}+B{}^\text{2}+2AB\cos \varphi }\],
where A and B are two forces, R is their resultant and \[\varphi \] is the angle between them.
Complete step by step answer:
Let the magnitude of force be = \[F\]
Given: \[R{}^\text{2}\text{ }=\text{ }3F{}^\text{2}\] where R is the resultant of 2 forces and \[F{}^\text{2}\] is the product of both the forces.
We have to find the angle between two forces.
Consider the angle between the equivalent forces to be \[\varphi \]
The magnitude of the resultant of the two forces is given by:
\[R=\sqrt{A{}^\text{2}+B{}^\text{2}+2AB\cos \varphi }\]
Here, \[A=B=F\]
therefore \[R=\sqrt{F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi }\]
Squaring both sides -
\[R{}^\text{2}=F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi \]
Now,
as \[R{}^\text{2}\text{ }=\text{ }3F{}^\text{2}\]
\[3F{}^\text{2}=F{}^\text{2}+F{}^\text{2}+2F{}^\text{2}\cos \varphi \]
or \[1=2\cos \varphi \]
Thus, \[\cos \varphi =0.5\]
\[\cos \varphi =\dfrac{1}{2}\]
\[\therefore\varphi ={{\cos }^{-1}}\dfrac{1}{2}={{60}^{\circ }}\]
Hence, the answer is option C.
Note:Even triangle law can be applied here, where the forces F and F acting at a point are shown based on magnitude and direction, by the two sides of a triangle taken in an order, and the obtained resultant is shown by the third side of the triangle which is taken in the opposite order. We have used parallelogram law where the two vectors acting at a point are shown based on magnitude and direction, by the two adjacent sides of a parallelogram generated from a particular point and the obtained resultant is represented by the diagonal of the parallelogram that is drawn from the same point.
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