Question

A body is under the action of two vectors (forces) 7N & 10N. Find the resultant of the two forces if the forces are inclined at angle of 60degrees to each other

Answer 1

Given

• A body is under the action of two vectors of 7N & 10N
• The Forces are inclined with an Angle of 60°

To Find

• The Resultant of these two forces

Solution

☯ Here we use the formula Resultant = √{F₁² + F₂² + 2F₁F₂ cosθ}

★ According to the given Question :

• F₁ = 7N
• F₂ = 10N
• θ = 60°

━━━━━━━━━━━━━━━━━━

Substituting the values

→ Resultant = √{F₁² + F₂² + 2F₁F₂ cosθ}

→ R = √{7² + 10² + 2(7)(10) × cos 60°}

→ R = √{49 + 100 + 140 × cos 60°}

❲ cos 60° = ½ ❳

→ R = √{49 + 100 + 140 × ½}

→ R = √{149 + 70}

→ R = √219

→ Resultant = 14.79 N

∴ The Resultant force will be 14.79 N

Answer 2

Answer:

$\huge \bf \: answer$

$\large { \boxed {\boxed { \sf{14.79 \: n}}}}$

$\huge \bf \: explanation$

$\sf √ \left({f1}^{2} + {f2}^{2} + 2f1f2 \times \cos \theta\right)$

$\sf \: \sqrt{}( {f1}^{2} + {f2}^{2} + 2(f1)(f2) \times60 \degree)$

$\sqrt{}({7}^{2} + {10}^{2} + 2(7)(10) \times 60)$

$\sqrt{} (49 + 100 + 140 \times 60)$

Cos 60⁰ = ½

$\sf \sqrt{} (49 + 100 + 140 \times \dfrac{1}{2} )$

$\sf \: \sqrt{}(149 + 70)$

$\sf \: \sqrt{} 219$

$\huge \bf \: resultant \: = 14.79 \: n$