Question

3
Find the magnitude of the two forces such that
if they act at a right angle and their resultant is √10N but if they act at 60° their resultant is √13​

Answer 1

Hey there!

Let the forces be F1 and F2 :

$\sqrt{F_1^2 + F_2^2} = \sqrt{10}$           ( As the angle between them is 90°)

When angle is 60°  :

$\sqrt{F_1^2 + F_2^2 + 2.F_1.F_2Cos60} = \sqrt{13}$  

$\sqrt{F_1^2+F_2^2+2. F_1.F_2.\frac{1}{2} } = \sqrt{13}$

$10+ F_1.F_2 = 13$

$F_1F_2 = 3$

Now ,

$(F_1+F_2)^2 = F_1^2+F_2^2 + 2F_1F_2$

                $= 10 + 2*3$

$F_1+F_2 = \sqrt{16}$     $= 4$         ------(1)    

Similarly ,

$(F_1-F_2)^2 = F_1^2 + F_2^2 -2F_1.F_2$

                 $= 10 - 2*3$

$F_1- F_2 = \sqrt{10-6}$    $= 2$      ------(2)

Now , let's add equation (1) and (2) :

$F_1 + F_2 =4 \\F_1 - F_2 = 2 \\------\\2F_1 = 6 \\------$

Hence, F1 = 6/2 = 3N

Also,

$F_1+F_2 = 4\\3 + F_2 = 4\\F_2 = 4-3 \\F_2 = 1$

Hence, F2 = 1 N          

∴ So, the magnitude of two forces is 3N and 1N