# Find the magnitude and the direction of the resultant force.Three forces in a plane act on an object. The forces are 7 N, 11 N, and 15 N. The angle between the 7 N and 11 N forces is 105°, between...

Find the magnitude and the direction of the resultant force.

Three forces in a plane act on an object. The forces are 7 N, 11 N, and 15 N. The angle between the 7 N and 11 N forces is 105°, between the 11 N and 15 N forces is 147°, and between the 15 N and 7 N forces is 108°.

I've been having a hard time trying to set up and solve for this problem.

Applications with Vectors

jeew-m | Certified Educator

Since we need to solve this equation in vector method we have to follow this procedure.

Let the unit vector to the X direction is i and unit vector to the Y direction is j.

Let’s assume that the force 11N is along the X axis and force 7N is at the left side of 11N.

Let angles measure to counter clock wise directions are positive.

Now you will get the forces like this.

Force 11N is along X axis

Force 7N is 105 degree counter clock wise to for 11N

Force 15N is 147 degree clock wise to the force 11N

Now if you use vector notation you can write the forces as;

Force 11N = 11i+0j = 11 i

Force 7N = 7*(cos105)i + 7*(cos(105-90))j = -1.8117 i + 6.7615 j

Force 15N = 15*(cos(360-147)i+15*(cos(360-147-90) j

=-12.5800 i -8.1696 j

So resultant force =11i - 1.8117 i + 6.7615 j - 12.5800 i - 8.1696 j

= -3.3917 i – 1.4081 j

Magnitude of the resultant force  = sqrt[(-3.3917)^2+(-1.4081)^2]

= 3.6724 N

Direction of the resultant force   = tan-1 (-1.4081/-3.3917)

= 22.546 deg

So the resultant force of 3.6724N is at 22.546 degrees counter clock wise angles to the 11N force.

najm1947 | Student

Let the direction of 7N force be the reference then, the dirction of 11N force is 105 deg and that of 15N force is 252 (105+147) deg, all angles measured counterclock wise from the direction of 7N force

Resolving the forces in direction of 7N force we get:

All forces along 7N force = 7 + 11cos(105) + 15cos(252) = -0.48N

All forces perpendicular to 7N force = 11sin(105) + 15sin(252) = -3.64N

The magnitude of the resultant force = sqrt{(-0.48)^2+(-3.64)^2) = 3.67N

Direction of the resultant force =Tan-1{(-3.64)/(-0.48)} = 262.5 degree in counterclock-wise direction of 7N force or 97.5 degree from the direction of 7N force in the clockwise direction

The magnitude of the resultant force is 3.67N at 97.5 degrees clockwise from direction of 7N force