MAGNITUDE OF ACCELERATION QUESTION!

Three forces in the x-y plane act on a 4.90 kg mass: 10.70 N directed at 77o, 7.80 N directed at 152o, and 3.50 N directed at 236o. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction. Calculate the magnitude of the acceleration.

Calculate the direction of the resultant force using the same sign convention as above (in degrees).

4.90 kg mass:
10.70 N directed at 77o,
7.80 N directed at 152o, and
3.50 N directed at 236o.

All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction. Calculate the magnitude of the acceleration.

Calculate the direction of the resultant force using the same sign convention as above (in degrees).

Resolving the forces in x – direction, force along x direction is
10.7 cos 77° + 7.8 cos 152° + 3.5 cos 236° = -6.437 N

Resolving the forces in y– direction, force along y direction is
10.7 sin 77° + 7.8 sin 152° + 3.5 sin 236° = 11.19 N

Net force is √ [-6.437² + 11.19²} = 12.91 N at angle (90 + α) where
α = tan inverse of (x/y) 6.437/ 11.19 = 29.90
(90 + α) = 119.9°

Magnitude of acceleration = force / mass = 12.91/ 4.9 = 2.65 m/s²

Direction is 119.9°measured from the positive x-axis, with positive angles in the Counter-Clockwise direction

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