Question

Find the magnitude and direction of two forces 40 N and 60 N
acting at a point with an included angle of 40° between them.
The 60N forces is horizontal.​

Answer 1

Answer:

R = 94.22 N

Alpha = 15.85⁰ or 15⁰51'

Answer 2

Resultant Force = 39.025 N

Angle of action = 15.83° from 60N force.

Let, F1 = Force acting horizontal = 60 N

F2 = Force acting at 40° = 40 N

Ф = angle subtended between them = 40°

Let R be the resultant force

We know,

R² = F1² + F2² + 2 × F1 × F2 × cos(180 - Ф)

∴ R² = 60²+40²+4800 × cos140

∴ R = √1522.98

∴ R = 39.025 N

And tan∅ = $\frac{F2*sin(si)}{F1+F2*cos(si)}$

∴ tan∅ = $\frac{40*sin40}{60+40*cos40}$

∴ ∅ = tan-1  $\frac{40*sin40}{60+40*cos40}$

∴ ∅ = 15.83