Three forces in the x-y plane act on a 8.10 kg mass: 11.70 N directed at 72o, 8.80 N directed at 131o, and 6.20 N directed at 211o. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction.Calculate the direction of the resultant force using the same sign convention as above (in degrees).
- JSAMLv 51 decade agoFavorite Answer
You need to break up all the forces into x and y-components and sum them up to get the resultant force (F_r).
Force 1 = 11.7 N, angle 72
x-component ---> 11.7*cos(72) = 3.62 N
y-component ---> 11.7*sin(72) = 11.13 N
Force 2 = 8.8 N, angle 131
x-component ---> 8.8*cos(131) = -5.78 N
y-component ---> 8.8*sin(131) = 6.64 N
Force 3 = 6.2 N, angle 211
x-component ---> 6.2*cos(211) = -5.31 N
y-component ---> 6.2*sin(211) = -3.19 N
F_rx = F_1x + F_2x + F_3x = 3.62-5.78-5.31 = -7.47 N
F_ry = F_1y + F_2y + F_3y = 11.13+6.64-3.19 = 14.58 N
Now find the magnitude and angle to find the resultant force:
|F_r| = sqrt[F_rx^2 + F_ry^2] = sqrt[(-7.47)^2+(14.58)^2] = 16.38 N
angle(F_r) = arctan(14.58/7.47) = 62.87 deg
Since the resultant force has a negative x-component and positive y-component, the force lies in Quadrant II. Thus, to get the exact angle using your notation:
angle = 180-62.87 = 117.13
Thus: resultant force = 16.38 N, angle 117.13
Hope this helps
- cfpopsLv 51 decade ago
The approach to this type of problem is to calculate the x and y components of each of separate forces, then sum all the x components to get a resultant x; then sum all of the y components to get a resultant y. Knowing x, y of the resultant force will allow you to calculate the hypoteneuse (magnitude of resulting force) and the angle with respect to the x-axis.
F1 = 11.7 N at 72 deg
F2 = 9.8 N at 131 deg
F3 = 6.2 N at 211 deg
F1x = -11.7/cos(72) = -37.86 N
F1y = -11.7/sin(72) = -11.17 N
F2x: 131 deg is 180-131 = 49 deg measured from -x axis
F2x = +8.8/cos(49) = +5.77 N
F2y = -8.8/sin(49) = -6.64 N
F3x: 211 deg is 211-180 = 31 deg measured from -x axis
F3x = +6.2/cos(31) = +5.31 N
F3y = +6.2/sin(31) = +3.19 N
Frx = F1x + F2x + F3x = -37.86 + 5.77 + 5.31 = -26.78 N
Fry = F1y + F2y + F3y = -11.17 - 6.64 + 3.19 = -14.62 N
Fa1: Resulting force angle with respect to -x axis:
Fa1 = acrtan(14.62/26.78) = 28.63 deg
Fa: Resulting force angle with respect to +x axis:
Fa = 180 + 28.63 = 208.63 degrees
Fr: Resulting force magnitude
Fr = 14.62/sin(28.63) = 14.62/0.479
Fr = 30.51 N
Resulting force is 30.51 N away from the mass at an angle of 208.63 degrees from the positive x-axis.
Hope this is correct, and hope it helps!
- 1 decade ago
Take the x-component of each force projected to the x-axis as Force(newtons) times cosine of the angle and add them together. Take the y-component as force times sine of the angle and add together the resultant vector [x,y] will have a magnitude of sqrt(x^2 + y^2) and angle arctan(y/x).