Question

two forces 6n and 8n are acting perpendicular to each other. calculate their resultant

Answer 1

Given :

Two Forces 6 N and 8 N are acting perpendicular to each other; so,

• F₁ = 6 N
• F₂ = 8 N
• Angle between forces, θ = 90°

To find :

• Resultant force of F₁ and F₂ acting perpendicularly , R = ?

Formula required :

• Formula to calculate magnitude of resultant vector

$\red{\clubsuit} \boxed{\sf{R^2=A^2+B^2+2\;A\;B\;cos\;\theta}}$

• Formula to calculate direction of resultant vector

$\red{\clubsuit} \boxed{\sf{tan\;\alpha=\dfrac{B\;sin\;\theta}{A+B\;cos\;\theta}}}$

[ where R is the magnitude of resultant of two vectors with magnitude A and B , θ is the angle between vectors A and B, and α is the angle between vector A and resultant vector R ]

Solution :

Using formula to calculate Magnitude of resultant vector

$\longmapsto\sf{R^2={F_1}^2+{F_2}^2+2\;F_1\;F_2\;cos\;\theta}$

$\longmapsto\sf{R^2={(6)}^2+{(8)}^2+2\;(6)\;(8)\;cos\;90^{\circ}}$

putting cos 90° = 0

$\longmapsto\sf{R^2=36+64+2\;(6)\;(8)(0)}$

$\longmapsto\sf{R^2=100}$

$\longmapsto\underline{\underline{\large{\red{\sf{R=10\;N}}}}}$

Using formula to calculate direction of resultant vector

Let, the angle between Force vector F₁ and resultant vector R be α

$\longmapsto\sf{tan\;\alpha=\dfrac{F_2 \;sin\;\theta}{F_1+F_2\;cos\;\theta}}$

$\longmapsto\sf{tan\;\alpha=\dfrac{(8)\;sin\;90^{\circ}}{(6)+(8)\;cos\;90^{\circ}}}$

putting sin 90° = 1 and cos 90° = 0

$\longmapsto\sf{tan\;\alpha=\dfrac{(8)(1)}{(6)+(8)(0)}}$

$\longmapsto\sf{tan\;\alpha=\dfrac{8}{6}=\dfrac{4}{3}}$

$\longmapsto\underline{\underline{\large{\red{\sf{\alpha=tan^{-1}\left(\dfrac{4}{3}\right)}}}}}$

• Therefore, Resultant vector has a magnitude of 10 Newtons and direction is at an angle $\bf{\alpha=tan^{-1}\left(\dfrac{4}{3}\right)}$ from the vector force F₁ .

Answer 2

Given ,

The two forces are 6 N and 8 N are acting perpendicular to each other

As we know that , the resultant between two vectors is given by

$\boxed{ \tt{R = \sqrt{ {(A)}^{2} + {(B)}^{2} + 2AB }}}$

Thus ,

$\tt \implies R = \sqrt{ {(6)}^{2} + {(8)}^{2} + 2 \times 6 \times 8 \times \cos(90) }$

$\tt \implies R = \sqrt{ 36 + 64 + 0}$

$\tt \implies R = \sqrt{100}$

$\tt \implies R = 10 \: \: newton$

The resultant of two given forces is 10 N