What is the angle between two forces F1=2N and F2=3N having resultant

1.	What is the angle between two forces F1=2N and F2=3N having resultant as 4N
Answer» GIVEN : Resultant of two forces = 4N F₁ = 2N F₂ = 3N To find : The angle between two forces. Solution : USING law of cosine that is, If two VECTORS A and B of magnitudes A and B are acting at an angle θ, then the MAGNITUDE of their resultant using parallelogram METHOD of vector addition is, $\boxed{ \sf R = \sqrt{ A^2 + B^2 + 2ABcos \theta}}$ By substituting the values, $\dashrightarrow\sf R = \sqrt{ F_1^{2} + F_2 ^{2} + 2ABcos \theta}$ $\dashrightarrow\sf 4 = \sqrt{2^{2} + 3 ^{2} + 2(2)(3)cos \theta}$ $\dashrightarrow\sf 16 = 4 + 9 + 12cos \theta$ $\dashrightarrow\sf 16 = 13 + 12cos \theta$ $\dashrightarrow\sf 16 - 13 = 12cos \theta$ $\dashrightarrow\sf 12cos \theta = 3$ $\dashrightarrow\sf cos \theta = \dfrac{3}{12}$ $\dashrightarrow\sf cos \theta = \dfrac{1}{4}$ $\dashrightarrow\sf \theta = 75 \degree32'$ Thus, the angle between two forces is 75° 32'

What is the angle between two forces F1=2N and F2=3N having resultant as 4N

Answer»

GIVEN :

Resultant of two forces = 4N

F₁ = 2N

F₂ = 3N

To find :

The angle between two forces.

Solution :

USING law of cosine that is,

If two VECTORS A and B of magnitudes A and B are acting at an angle θ, then the MAGNITUDE of their resultant using parallelogram METHOD of vector addition is,

$\boxed{ \sf R = \sqrt{ A^2 + B^2 + 2ABcos \theta}}$