Question

Find the magnitude of the resultant of the concurrent forces of 8 N, 12 N, 15N and 20 N making angles of 30°, 70°, 120°and 155° respectively with a fixed-line.​

Answer 1

Explanation:

The resultant of two vectors:

P and Q is given by

$r = \sqrt{p2 + q2 + 2pq \cos(θ) }$

where, P and Q are magnitudes of two vectors and θ is the angle between two vectors.

Here θ=? P=8N Q=15N R=17N

Therefore,

$17 = \sqrt{ {8}^{2} + {15}^{2} + 2 \times 8 \times 15 \ \cos(θ) }$

289= 64 + 225 + 240 cosθ

289= 289 + 240 cosθ

289 - 289 = 240 cosθ

cosθ = 0

θ=90⁰

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Answer 2

Answer:

Explanation:

The resultant of two vectors:

P and Q is given by

r = \sqrt{p2 + q2 + 2pq \cos(θ) } r=

p2+q2+2pqcos(θ)

where, P and Q are magnitudes of two vectors and θ is the angle between two vectors.

Here θ=? P=8N Q=15N R=17N

Therefore,

17 = \sqrt{ {8}^{2} + {15}^{2} + 2 \times 8 \times 15 \ \cos(θ) } 17=

8

2

+15

2

+2×8×15 cos(θ)

289= 64 + 225 + 240 cosθ

289= 289 + 240 cosθ

289 - 289 = 240 cosθ

cosθ = 0

θ=90⁰