Question

What is the angle between two forces of 2N and 3N having resultant as 4N? CLASS - XI PHYSICS (Kinematics) mention properly i am confusing

Answer 1

Let the two vectors be A and B, and their resultant be R.

Let the angle between A and B be θ

A = 2 N

B = 3 N

R = 4 N

Now R² = A² + B² + 2AB cos θ

So, 4² = 2² + 3² + 2×2×3 cos θ

So, 16 = 4 + 9 + 12 cos θ

So, 16 = 13 + 12 cos θ

So, 3 = 12 cos θ

So, 3/12 = cos θ

So, cos θ = 1/4

$\implies \sf \theta = cos^{-1} \: \left( \dfrac{1}{4} \right)$

Inverse cosine is also called arc cosine , or simply acos

So, θ = acos (1/4)

Answer 2

Answer:

Explanation: A=2N

B=3N and C= 4N

R²= A²+ B²+2ABcos∅

4²= 2²+3²+2×2×3 cos∅

16= 4+9+12 cos∅

16= 13+12 cos∅

16-13= 12 cos∅

12 cos∅=3

4 cos∅=1

so finally, cos∅=1/4

and ∅=cos^-1 (1/4)