Question

if the resultatant of two forces of magnitude p and 2p is perpendicular to pthen the angle between the forces is

Answer 1

Answer:

The angle between the forces is 120°.

Explanation:

Given that,

First vector = P

Second vector = 2P

Let R be the magnitude of resultant.

Let Ф be the angle between R and 2P.

Using cosine rule

The magnitude of resultant P and 2P is

R=2p\cos\theta

Now,

R^2=P^2+(2P)^2+2\times P\times 2P\cos(90+\theta)

4P^2\cos^2\theta=P^2+4P^2+4P^2-\sin\theta

4\cos^2\theta=5-4\sin\theta

4(1-\sin^2\theta)=5-4\sin\theta

4\sin^2\theta-4\sin\theta+1=0

(2\sin\theta-1)^2=0

2\sin\theta=1

\theta = sin^{-1}\dfrac{1}{2}

\theta=\dfrac{\pi}{6}=30^{\circ]

The angle between the forces is

\alpha=90^{\circ}+\theta

\alpha=90+30

\alpha=120^{\circ}

Hence, The angle between the forces is 120°.

Answer 2

The angle between the forces is 120°.

Explanation:

Given that,

First vector = P

Second vector = 2P

Let R be the magnitude of resultant.

Let Ф be the angle between R and 2P.

Using cosine rule

The magnitude of resultant P and 2P is

R=2p\cos\thetaR=2pcosθ

Now,

R^2=P^2+(2P)^2+2\times P\times 2P\cos(90+\theta)R2=P2+(2P)2+2×P×2Pcos(90+θ)

4P^2\cos^2\theta=P^2+4P^2+4P^2-\sin\theta4P2cos2θ=P2+4P2+4P2−sinθ

4\cos^2\theta=5-4\sin\theta4cos2θ=5−4sinθ

4(1-\sin^2\theta)=5-4\sin\theta4(1−sin2θ)=5−4sinθ

4\sin^2\theta-4\sin\theta+1=04sin2θ−4sinθ+1=0

(2\sin\theta-1)^2=0(2sinθ−1)2=0

2\sin\theta=12sinθ=1

\theta = sin^{-1}\dfrac{1}{2}θ=sin−121

\theta=\dfrac{\pi}{6}=30^{\circ]

The angle between the forces is

\alpha=90^{\circ}+\thetaα=90∘+θ

\alpha=90+30α=90+30

\alpha=120^{\circ}α=120∘

Hence, The angle between the forces is 120°.

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