Two forces of magnetude 6N and 10N act on a particle.their resultant is of magnitude 12N.find the angle between the two forces.
Make a rough sketch of the two vectors of 6N and 10N, completing a parallelogram so that the major diagonal is 12N
So looking at the triangle with sides 10,6, and 12
we can find the obtuse angle Ø by
12^2 = 6^2 + 10^2 - 2(6)(12)cosØ
144 cosØ = -8
cosØ = -1/18
Ø = appr 93.18°
then we can find the angle opposite side of length 6
sinA/6 = sin93.18/12
sinA = .499..
angle A = 29.95
leaving 56.87° for the third angle
so the angle between them is 56.87° + 29.95°
or 86.82°
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To find the angle between two forces, we can use the cosine formula. The cosine of the angle between the two forces is equal to the dot product of the forces divided by the product of their magnitudes.
Let's denote the magnitude of the first force as F1 = 6N and the magnitude of the second force as F2 = 10N. We are given that the resultant force has a magnitude of 12N.
Using the cosine formula, we have:
cosθ = (F1 * F2) / (|F1| * |F2|)
= (6N * 10N) / (|6N| * |10N|)
= 60N^2 / (6N * 10N)
= 60 / 60
= 1
Therefore, cosθ = 1.
To find the angle θ, we need to find the inverse cosine (also known as arccos) of 1. The inverse cosine of 1 is 0° in degrees or 0 radians.
So, the angle between the two forces is 0° or 0 radians.