Question

Two forces 20N and 25N
act on an
object if the
resultant force is 39 N
find angle between the force doing trigonometry method​

Answer 1

Given :

• Two forces of magnitude 20N and 25N are acting on an object
• Resultant of Forces is given as 39 N

To Find :

• Angle between the Forces

Solution :

Two forces are of magnitude 20N and 25N respectively. Let's suppose that

$\: \: \: \: \: \: \: \: \bullet \: \sf{F_1 \: = \: 20 \: N}$

$\: \: \: \: \: \: \: \: \bullet \: \sf{F_2 \: = \: 25 \: N}$

And,

$\: \: \: \: \: \: \: \: \bullet \: \sf{F_{net} \: = \: 39 \: N}$

Use formula for parallelogram law of Vector addition :

$\implies \sf{F_{net} \: = \: \sqrt{F_1^2 \: + \: F_2 ^2 \: + \: 2 F_1 F_2 \cos \theta}} \\ \\ \implies \sf{39 \: = \: \sqrt{20^2 \: + \: 25^2 \: + \: 2 \: \times \: 20 \: \times \: 25 \: \times \: \cos \theta}}$

Square both sides

$\implies \sf{39^2 \: = \: 20^2 \: + \: 25^2 \: + \: 40 \: \times \: 25 \: \times \: \cos \theta} \\ \\ \implies \sf{39^2 \: = \: 20^2 \: + \: 25^2 \: + \: 1000 \cos \theta} \\ \\ \implies \sf{1521 \: = \: 400 \: + \: 625 \: + \: 1000 \: \cos \theta} \\ \\ \implies \sf{1521 \: = \: 1025 \: + \: 1000 \cos \theta} \\ \\ \implies \sf{1000 \cos \theta \: = \: 1521 \: - \: 1025} \\ \\ \implies \sf{1000 \cos \theta \: = \: 496} \\ \\ \implies \sf{\cos \theta \: = \: \dfrac{496}{1000}} \\ \\ \implies \sf{\cos \theta \: = \: 0.496} \\ \\ \implies \sf{\cos \theta \: \approx \: 0.5} \\ \\ \implies \sf{\theta \: = \: cos^{-1} ( 0.5)} \\ \\ \implies \sf{\theta \: = \: 60^{\circ}}$