Question

Three forces in the x-y plane act on a 6.70 kg mass: 16.30 N directed at 72°, 14.00 N directed at 128°, and 12.20 N directed at 199°. 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).

Answer 1

Horizontal componet of first force: F_1x = F₁ cosθ₁ = 5.03 N
Vertical component of first force: F_1y = F₁ sinθ₁ = 15.5 N

Horizontal componet of second force:F_2x = F₂ cos θ₂ = -8.619 N
Vertical component of second force:F_2y = F₂ sinθ₂ = 11.03 N

Similarly for F3.

F_3x = -11.53 N
F_3y = -3.97 N

Then add up the components to get total Fx and Fy.

Fx = -15.11 N

Fy = 22.56 N

Use Pythagorus to get the magnitude of the force.
F = sqrt (Fx^2 + Fy^2) = 27.15 N

To get acceleration using Newton's 2nd Law:
a = F/m = 4.05 m/s²

As a bonus, if you cared about the direction of the force/acceleration:
direction = arctangent (Fy/Fx) = -88.45 degree

180 - 88.45 = 91.55 degree.