Question

Three forces in the x-y plane act on a 4.80 kg mass: 9.90 N directed at 502.0°, 10.20 N directed at 135.0°, and 5.90 N directed at 205.0°. All angles are measured from the positive x-axis, with positive angles in the counter-clockwise direction. Calculate the magnitude of the acceleration.

Weight= 47N.

I know I'm supposed to start by getting the x and y components of the each vectors. Which I found to be.

F1x= 9.90 cos 45
F2x= 10.20 cos 38
F3x 5.90 cos 25

F1y= 9.90 sin 45
F2y= 10.20 sin 38
F3y= 5.90 sin 25

Then I'm supposed to use the Second Law : ΣF=m*a ; a=ΣF/m

I'm not really sure how/where to plug those xy components in.

Answer 1

add the three forces together component-wise

Answer 2

So. Sum of x components = 20.39 . Sum of y components = 15.77. What's next?

Answer 3

now find the magnitude of this resultant force,

Answer 4

25.78 ? sqrt(20.39^2+15.77^2)

Answer 5

dow divide by mass?

Answer 6

now*