What is the angle between two vector forces of equal magnitude such that the resultant is one-third as
much as either of the original forces
Answers
Answer:
The angle between two vectors = 160.81°
Explanation:
As per the question,
Resultant is given by:
Resultant=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos\theta}Resultant=
F
1
2
+F
2
2
+2F
1
F
2
cosθ
Now,
Its given that two forces are equal in magnitude.
∴ F₁ = F₂ = F
Also, Resultant force is one third of the original forces.
R =\frac{F}{3}R=
3
F
Therefore,
Resultant=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos\theta}Resultant=
F
1
2
+F
2
2
+2F
1
F
2
cosθ
On putting all the values , we get
\frac{F}{3}=\sqrt{F^{2}+F^{2}+2F^{2}cos\theta}
3
F
=
F
2
+F
2
+2F
2
cosθ
On squaring both sides, we get
$$\frac{F^{2}}{9}=2F^{2}+2F^{2}cos\theta}$$
$$\begin{lgathered}1+cos\theta=\frac{1}{18}=0.0555\\cos\theta=-0.944\\\theta =cos^{-1}(-0.944)\end{lgathered}$$
θ = 160.81°
Hence, the angle between two vectors = 160.81°
