Question

Two forces 20N and 40N act simultaneously at an
angle 45° to each other. Find the magnitude and
directions of their resultant?
olution
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Answer 1

Given

• Force₁ = 20 N
• Force₂ = 40 N
• They are inclined at an angle of 45°

To Find

• Magnitude and direction of their resultant

Solution

● Resultant = √{A²+B²+2AB cosθ}

✭ Resultant :

→ Resultant = √{A²+B²+2AB cosθ}

• A = 20 N
• B = 40 N
• θ = 45°

→ R = √20² + 40² + 2(20)(40)

→ R = √400 + 1600 + 1600

→ R = √3600

→ R = 60 N

✭ Direction

→ cos ω = 40/60

→ cos ω = 0.67

→ ω = cos⁻¹ 0.67

∴ The magnitude and the direction of their resultant is cos⁻¹ 0.67

Answer 2

Answer :-

• Magnitude of resultant (R) = $\boxed{\tt{\red{55.96 \:N}}}$

• Direction of resultant (ϕ) = $\boxed{\tt{\red{30.4°}}}$

Given :-

• Force (A) = 20N

• Force (B) = 40N

• Angle (θ) = 45°

To find :-

• Magnitude and direction of resultant

Formula used :-

• $\boxed{\sf{R=\sqrt{A²+B²+2ABcosθ}}}$

• $\boxed{\sf{tanϕ = \dfrac{Bsinθ}{A + Bcosθ}}}$

Solution :-

$\sf{➪\:R=\sqrt{A²+B²+2ABcosθ}}$

$\sf{➪\:R=\sqrt{(20)²+(40)²+2(20)(40)cos45°}}$

$\sf{➪\:R=\sqrt{2000+1600\times \dfrac{1}{\sqrt{2}}}}$

$\sf{➪\:R=\sqrt{2000+800\times \sqrt{2}}}$

$\sf{➪\:R=\sqrt{3131} = }$ $\bold{\red{55.96 \:N}}$

$\:$

$\sf{➪\:tanϕ = \dfrac{Bsinθ}{A + Bcosθ}}$

$\sf{➪\:tanϕ = \dfrac{40sin45°}{20 + 40cos45°}}$

$\sf{➪\:tanϕ = \dfrac{\dfrac{40}{\cancel{\sqrt{2}}}}{\dfrac{20\sqrt{2} + 40}{\cancel{\sqrt{2}}}}}$

$\sf{➪\:tanϕ = \dfrac{2}{\sqrt{2} +2}}$

$\sf{➪\:ϕ = tan^{-1} (0.58)}$

$\sf{➪\:ϕ = }$ $\sf{\red{30.4°}}$

Use this site for checking the answer:

https://www.easycalculation.com/physics/classical-physics/resultant-vector.php