A force \[F\] makes an angle \[20^\circ \] with another\[F\] . The resultant of two forces is:
A. \[F\cos 20^\circ \]
B. \[2F\cos 20^\circ \]
C. \[2F\cos 10^\circ \]
D. \[F\cos 10^\circ \]
Answer
615k+ views
Hint: First we find an expression, which gives the resultant of two vector quantities, when two forces are inclined by an angle.
The expression includes the trigonometric ratio cosine.
Complete step by step answer:
In the given problem,
The two forces, which are equal in magnitude i.e. identical in nature, are \[F\] .
The two forces are inclined to each other by an angle of \[20^\circ \] .
We are asked to find the resultant of the two forces.
We have a formula which gives the resultant of two vector quantities, which are inclined by an angle \[\theta \] :
\[R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \] …… (1)
Where,
\[R\] indicates the resultant of the two quantities.
\[P\] indicates one vector quantity.
\[Q\] indicates another vector quantity.
\[\theta \] indicates the angle by which the two quantities are inclined to each other.
According to given question,
\[P = F\]
\[Q = F\] and
\[\theta = 20^\circ \]
Substituting these values in the equation (1), we get:
$ R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \\ $
$ \implies R = \sqrt {{F^2} + {F^2} + 2 \times F \times F\cos 20^\circ } \\ $
$ \implies R = \sqrt {2{F^2} + 2{F^2}\cos 20^\circ } \\ $
$ \implies R = \sqrt {2{F^2}\left[ {1 + \cos \left( {2 \times 10^\circ } \right)} \right]} \\ $
To simplify the above equation further, we have a trigonometric formula, which is:
\[1 + \cos 2\alpha = 2{\cos ^2}\alpha \]
Here, in this numerical, \[\alpha = 10^\circ \]
Again, manipulating the equation of resultant, we have,
$ \implies R = \sqrt {2{F^2} \times 2{{\cos }^2}10^\circ } \\ $
$ \implies R = \sqrt {4{F^2}{{\cos }^2}10^\circ } \\ $
$ \implies R = 2F\cos 10^\circ \\ $
Hence, the resultant of the two forces is found to be \[2F\cos 10^\circ \] .
So, the correct answer is “Option C”.
Additional Information:
Vector, a quantity which has both magnitude and direction in physics. An arrow whose direction is the same as that of the quantity and whose length is equal to the magnitude of the quantity is usually represented. If there is magnitude and path for a vector, it does not have position.
Note:
Note: It is important to note that the angle between two vector quantities affects the magnitude of the resultant. The resultant between two vectors is maximum when the angle between them is \[0^\circ \] whereas it is minimum when the angle between them is \[180^\circ \] .
The expression includes the trigonometric ratio cosine.
Complete step by step answer:
In the given problem,
The two forces, which are equal in magnitude i.e. identical in nature, are \[F\] .
The two forces are inclined to each other by an angle of \[20^\circ \] .
We are asked to find the resultant of the two forces.
We have a formula which gives the resultant of two vector quantities, which are inclined by an angle \[\theta \] :
\[R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \] …… (1)
Where,
\[R\] indicates the resultant of the two quantities.
\[P\] indicates one vector quantity.
\[Q\] indicates another vector quantity.
\[\theta \] indicates the angle by which the two quantities are inclined to each other.
According to given question,
\[P = F\]
\[Q = F\] and
\[\theta = 20^\circ \]
Substituting these values in the equation (1), we get:
$ R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \\ $
$ \implies R = \sqrt {{F^2} + {F^2} + 2 \times F \times F\cos 20^\circ } \\ $
$ \implies R = \sqrt {2{F^2} + 2{F^2}\cos 20^\circ } \\ $
$ \implies R = \sqrt {2{F^2}\left[ {1 + \cos \left( {2 \times 10^\circ } \right)} \right]} \\ $
To simplify the above equation further, we have a trigonometric formula, which is:
\[1 + \cos 2\alpha = 2{\cos ^2}\alpha \]
Here, in this numerical, \[\alpha = 10^\circ \]
Again, manipulating the equation of resultant, we have,
$ \implies R = \sqrt {2{F^2} \times 2{{\cos }^2}10^\circ } \\ $
$ \implies R = \sqrt {4{F^2}{{\cos }^2}10^\circ } \\ $
$ \implies R = 2F\cos 10^\circ \\ $
Hence, the resultant of the two forces is found to be \[2F\cos 10^\circ \] .
So, the correct answer is “Option C”.
Additional Information:
Vector, a quantity which has both magnitude and direction in physics. An arrow whose direction is the same as that of the quantity and whose length is equal to the magnitude of the quantity is usually represented. If there is magnitude and path for a vector, it does not have position.
Note:
Note: It is important to note that the angle between two vector quantities affects the magnitude of the resultant. The resultant between two vectors is maximum when the angle between them is \[0^\circ \] whereas it is minimum when the angle between them is \[180^\circ \] .
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

Name the Largest and the Smallest Cell in the Human Body ?

