Question

5. Calculate the angle between a 2 N force and a 3 N
force so that their resultant is 4 N. (Ans. 75°31')

Answer 1

Answer:

• The angle (θ) b/w them is 75.52°

Given:

1. Resultant Force (R) = 4 N.
2. Two forces are 2 N & 3 N.

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bigstar\;\underline{\boxed{\sf R^{2}=F_{1}^{2}+F_{2}^{2}+2\; F_{1}\;F_{2}\cos \theta}}$

Here,

• R Denotes Resultant.
• F₁ & F₂ Denotes Forces.

Substituting the values,

$\longrightarrow\sf (4)^{2}=(2)^{2}+(3)^{2}+2\times 2\times 3\cos \theta \\\\\\\longrightarrow\sf 16 = 4+9+12\cos \theta\\\\\\\longrightarrow\sf 16=13+12\cos \theta\\\\\\\longrightarrow\sf 12\cos \theta =16 - 13\\\\\\\longrightarrow\sf 12\cos \theta=3\\\\\\\longrightarrow\sf \cos \theta =\dfrac{3}{12}\\\\\\\longrightarrow\sf \cos\theta =\dfrac{1}{4}\\\\\\\longrightarrow\sf \theta =cos^{-1}\Bigg\lgroup\dfrac{1}{4}\Bigg\rgroup$

$\longrightarrow\sf \theta = 75.52^{\circ}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \theta = 75.52^{\circ}}}}}$

∴ The angle (θ) b/w them is 75.52°.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathbb\red{ANSWER:-}$

The angle (θ) b/w them is 75.52°