add a force of 8N directed northwards to a force of 6N directed westwardsand .Calculate the magnitude and the direction of resultant with north

Let the resultant be R. Let the resultant make a angle A with the North.

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Let the force northwards be A with 8N magnitude. and the force westwards be B with 6N magnitude, and their resultant be R.

Angle between A and B is 90 degrees.

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Therefore by the formula as below:-

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R^2 = A^2 + B^2 + 2AB (Costheta)

R^2 = (8)^2 + (6)^2 + 2(8)(6) Cos90

R^2 = 64 + 36 + 96 (0)

R^2 = 100

R = 10 N

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Second part of question says to find direction of resultant from north. (i.e) Angle between R and A forces.

Now here for resultant of R and A forces is B .

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Hence by the same formula:-

B^2 = R^2 + A^2 + 2AR (cos theta) .......... [ To find :- Theta ]

(6)^2 = (10)^2 + (8)^2 + 2(10)(8) (Cos theta)

36 = 100 + 64 + 160 ( Costheta)

Cos theta = (36 - 100 - 64) / 160 = -128 / 160 = -4/5

Theta = Cos inv of (-4/5) from the A force vector.

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