Question

8.Two forces of magnitude 8 N and 15 N respectively
act at a point. If the resultant force is 17 N, the
angle between the forces has to be
(1) 60°
(2) 45°
(3) 90°
(4) 30​

Answer 1

$\huge\bigstar\sf{\underline{Given\::-}}$

• $\sf{\ F_{1}=8N}$

• $\sf{\ F_{2}=15N}$

• $\sf{\ F_{resultant}=17N}$

$\huge\bigstar\sf{\underline{To\:Find\::-}}$

• Angle between Force $\sf{\ F_{1}}$ and $\sf{\ F_{2}}$

$\huge\bigstar\sf{\underline{Solution\::-}}$

• Let the $\sf\ F_{resultant}$ be the resultant of these two forces.

$\sf\small{\red{\ F_{resultant}=\sqrt{\ (\ F_{1})^{2}+\ (\ F_{2})^{2}+2\ F_{1}\ F_{2} Cos(\theta)}}}$

${}$

On squaring both side

${}$

$\sf\small{\ (\ F_{resultant})^{2}=\ (\ F_{1})^{2}+\ (\ F_{2})^{2}+2\ F_{1}\ F_{2} Cos(\theta)}$

$\mapsto\sf{\ (17)^{2}=\ (8)^{2}+\ (15)^{2}+2(8)(15) Cos(\theta)}$

$\mapsto\sf{289=64+225+240 Cos(\theta)}$

$\mapsto\sf{289=289+240 Cos(\theta)}$

$\mapsto\sf{240 Cos(\theta)=289-289}$

$\mapsto\sf{240 Cos(\theta)=0}$

$\mapsto\sf{Cos(\theta)=0}$

$\mapsto\sf{ Cos{\theta}= Cos{\dfrac{\pi}{2}}}$

$\mapsto\sf{\cancel{Cos}{\theta}=\cancel{Cos}{\dfrac{\pi}{2}}}$

$\mapsto\sf{\theta=\dfrac{\pi}{2}}$

The angle is in radian we can change into degree by multiplying it with $\sf{\dfrac{180}{\pi}}$

$\sf{\theta=\dfrac{\pi}{2}\times \dfrac{180}{\pi}}$

$\sf{\theta=\dfrac{\cancel{\pi}}{\cancel{2}}\times \dfrac{\cancel{180}}{\cancel{\pi}}}$

$\sf{\boxed{\boxed{\pink{\dag{\theta = 90°}}}}}$

• So the angle between these two forces is 90° , means these forces are perpendicular to each other .

Answer 2

90°

Given:-

$\;\;\;\bullet\;\sf F_1 = 8N \\ \\ \;\;\;\bullet\;\sf F_2 = 15N \\ \\ \;\;\;\sf F_{resultant} = 17N$

To find:-

$\;\;\;\sf Angle\;b/w\;F_1 and F_2$

Solution:-

As we know that,

If $\sf F_{resultant}$ is the resultant of these two forces $\sf F_1 and F_2$

$\dag\;{\underline{\boxed{\bf{\blue{ F_{resultant} = \sqrt{( F_1 )^2 + ( F_2 )^2 + 2 F_1 F_2 \cos{ \theta}}}}}}}$

★ Squaring both sides:-

$:\implies\sf ( F_resultant )^2 = ( F_1 )^2 + ( F_2 )^2 + 2 F_1 F_2 \cos{ \theta} \\ \\ \dashrightarrow\sf (17)^2 = (8)^2 + (15)^2 + 2 \times 8 \times 15 \times \cos{ \theta} \\ \\ \dashrightarrow\sf 289 = 64 + 225 + 240 \cos{ \theta} \\ \\ \dashrightarrow\sf 240 \cos{ \theta} = 289 - 289 \\ \\ \dashrightarrow\sf 240 \cos{ \theta} = 0 \\ \\ \dashrightarrow\sf \cos{ \theta} = 0$

As we know that,

Value of $\theta$ is 0 at cos90°

Therefore,

$\dashrightarrow\sf \cos{ \theta} = \cos{90^\circ}\qquad\bigg\lgroup\bf \cos{90^\circ} = 0\bigg\rgroup \\ \\ \dashrightarrow\sf { \cancel{ \cos}}{ \theta} = { \cancel{ \cos}}{90^\circ} \\ \\ \dag\;{\underline{\boxed{\bf{\red{\dashrightarrow \theta = 90^\circ}}}}}$

★ Therefore, the angle b/w these two forces is 90°

★ So, we can say that these forces are perpendicular to each other.

$\dag$ Hence, Option (3) is correct.

$\rule{200}{3}$