Determine the resultants for the following vector addition. 125 kg m/s 25 degree Eof N + 80 kg m/s 48 degree Sof W

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The resultant of the forces, A = 125 kg*m/s 25 degree E of N and B = 80 kg*m/s 48 degree S of W. Has to be determined.

Take the components of each of the forces along the E-W and N-S direction.

Force A is equivalent to 125*cos 25 towards the North and 125*sin 25 towards the East. Force B is equivalent to 80*cos 48 towards the West and 80*sin 48 towards the South.

Adding the two forces gives 125*cos 25 - 80*sin 48 `~~` 53.68 N towards the North and 80*cos 48 - 125*sin 25 `~~` 0.70 N towards the West. the magnitude of the resultant force is sqrt(53.68^2 + 0.7^2) = 53.68 N. The direction of the force is 0.747 degrees West of North.

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