
The resultant of two forces, one double the another in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is?
Answer
552.3k+ views
Hint: Force is a vector and it follows vector’s law of addition. There are two laws of adding vectors which are mostly used. These are Triangles law of vector addition and parallelogram’s law of vector addition.
Complete step by step answer:
Let us assume that the two forces are \[{{F}_{1}}\And {{F}_{2}}\]. Let F be the resultant of the two forces. It is given that the magnitude of one force is two times a second. Writing this mathematically,
\[{{F}_{1}}=2{{F}_{2}}\]
Also, the angle between the resultant force and the smaller force is \[90{}^\circ \]
Using the law of vector addition, \[\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}=F\]
Now taking the dot product of F with the smaller force. Since both are perpendicular so the dot product comes out to be zero.
\[\begin{align}
& \overrightarrow{{{F}_{1}}}.\overrightarrow{{{F}_{2}}}=0 \\
& (\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}).\overrightarrow{{{F}_{2}}}=0 \\
& \overrightarrow{{{F}_{1}}}.\overrightarrow{{{F}_{2}}}+\overrightarrow{F_{2}^{2}}=0 \\
\end{align}\]
\[{{F}_{1}}{{F}_{2}}\operatorname{Cos}\theta +F_{2}^{2}=0\]
Using \[{{F}_{1}}=2{{F}_{2}}\]
\[\begin{align}
& 2{{F}_{2}}^{2}\cos \theta +{{F}_{2}}^{2}=0 \\
& 2\cos \theta +1=0 \\
& \cos \theta =-0.5 \\
\end{align}\]
Negative for cos is in second and third quadrant which gives \[\theta =\dfrac{2\pi }{3}\]
Additional Information:
The two vectors are said to be perpendicular when the angle between the vectors is \[90{}^\circ \] and the two vectors are said to be parallel to each other if the angle between them is \[0{}^\circ \]
Also, while adding or subtracting the two or more vectors we have to keep in mind what is the angle between them. This explains the vector law of addition.
Note:
This is an example of how to use the vector addition and what are the special cases when the vectors are perpendicular to each other or parallel to each other.
Complete step by step answer:
Let us assume that the two forces are \[{{F}_{1}}\And {{F}_{2}}\]. Let F be the resultant of the two forces. It is given that the magnitude of one force is two times a second. Writing this mathematically,
\[{{F}_{1}}=2{{F}_{2}}\]
Also, the angle between the resultant force and the smaller force is \[90{}^\circ \]
Using the law of vector addition, \[\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}=F\]
Now taking the dot product of F with the smaller force. Since both are perpendicular so the dot product comes out to be zero.
\[\begin{align}
& \overrightarrow{{{F}_{1}}}.\overrightarrow{{{F}_{2}}}=0 \\
& (\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}).\overrightarrow{{{F}_{2}}}=0 \\
& \overrightarrow{{{F}_{1}}}.\overrightarrow{{{F}_{2}}}+\overrightarrow{F_{2}^{2}}=0 \\
\end{align}\]
\[{{F}_{1}}{{F}_{2}}\operatorname{Cos}\theta +F_{2}^{2}=0\]
Using \[{{F}_{1}}=2{{F}_{2}}\]
\[\begin{align}
& 2{{F}_{2}}^{2}\cos \theta +{{F}_{2}}^{2}=0 \\
& 2\cos \theta +1=0 \\
& \cos \theta =-0.5 \\
\end{align}\]
Negative for cos is in second and third quadrant which gives \[\theta =\dfrac{2\pi }{3}\]
Additional Information:
The two vectors are said to be perpendicular when the angle between the vectors is \[90{}^\circ \] and the two vectors are said to be parallel to each other if the angle between them is \[0{}^\circ \]
Also, while adding or subtracting the two or more vectors we have to keep in mind what is the angle between them. This explains the vector law of addition.
Note:
This is an example of how to use the vector addition and what are the special cases when the vectors are perpendicular to each other or parallel to each other.
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