Question

Two forces acting on a particle in opposite directions have the resultant of 10N. If they act at right angles to each other, the resultant it 60N. Fing the 2 forces.

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

Case : 1

Angle b/w two force vectors = 180°

Resultant force = 10N

Case : 2

Angle b/w two force vectors = 90°

Resultant force = 60N

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

We have to find magnitude of forces.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

◈ By triangle law or parallelogram law of vector addition, the magnitude of resultant of resultant R at two vectors P and Q inclined to each other at angle θ, is given by

$\bigstar\:\underline{\boxed{\bf{\red{R=\sqrt{P^2+Q^2+2PQ\cos\theta}}}}}$

✪ Case : 1

$:\implies\sf\:10=\sqrt{{F_1}^2+{F_2}^2+2F_1F_2\cos180\degree}$

$:\implies\sf\:10=\sqrt{{F_1}^2+{F_2}^2-2F_1F_2}$

$:\implies\sf\:10=\sqrt{(F_1-F_2)^2}$

$:\implies\bf\:F_1-F_2=10\:\longrightarrow\:(I)$

✪ Case : 2

$:\implies\sf\:60=\sqrt{{F_1}^2+{F_2}^2+2F_1F_2\cos60\degree}$

$:\implies\sf\:60=\sqrt{{F_1}^2+{F_2}^2+2F_1F_2}$

$:\implies\sf\:10=\sqrt{(F_1+F_2)^2}$

$:\implies\bf\:F_1+F_2=60\:\longrightarrow\:(II)$

By solving both equations, we get

• $\bf{F_1=35N}$
• $\bf{F_2=25N}$