Question
Two force 1oN and 20N are inclined at angle 60degree . find the magnitude and direction of the force
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To find the magnitude of the resulting force, we can use the Law of Cosines:
F^2 = (10N)^2 + (20N)^2 - 2(10N)(20N)cos(60°)
F^2 = 100N^2 + 400N^2 - 400N^2
F = 20N
So the magnitude of the resulting force is 20N.
To find the direction of the force, we can use the Law of Sines:
sin(60°) / F = sin(A) / 10N
sin(A) = (10N sin(60°)) / F
sin(A) = (10N sin(60°)) / 20N
sin(A) = 0.5
A = 30.0°
So the direction of the force is 30° above the 10N force (as measured from the positive x-axis).
F^2 = (10N)^2 + (20N)^2 - 2(10N)(20N)cos(60°)
F^2 = 100N^2 + 400N^2 - 400N^2
F = 20N
So the magnitude of the resulting force is 20N.
To find the direction of the force, we can use the Law of Sines:
sin(60°) / F = sin(A) / 10N
sin(A) = (10N sin(60°)) / F
sin(A) = (10N sin(60°)) / 20N
sin(A) = 0.5
A = 30.0°
So the direction of the force is 30° above the 10N force (as measured from the positive x-axis).