Respuesta :
If you add these vectors, the magnitude of the resultant vector will be
|R| = ( (80)^2 + (100)^2 +2(80)(100)cos60)^(1/2) = 156.205
Hence the magnitude of the single force that can be used in place of these two forces is 156.205 N.
Also, the magnitude of the force which will balance these two is 156.205 N acting exactly opposite to the resultant of these two forces.
If you are uncomfortable with vector sum, you can solve this by resolving the vectors.
place the forces on the cartesian plane in such a way that the 80N force lies in first quadrant and the 100N force lies in the fourth quadrant (each force making 30° with x axis), X axis being the angle bisector of the angle b/w the two forces.
Now you can resolve them into x and y components
for 80N, Fx = 80cos30, Fy = 80sin30
for 100N, Fx = 100cos30, Fy = 100sin30
Clearly you can see the resultant is
Fx = 180cos30
Fy = 20sin30
And magnitude of this force is = ((180cos30)^2 + (20sin30)^2)^1/2 = 156.205
If you need direction, calculate Fy/Fx
|R| = ( (80)^2 + (100)^2 +2(80)(100)cos60)^(1/2) = 156.205
Hence the magnitude of the single force that can be used in place of these two forces is 156.205 N.
Also, the magnitude of the force which will balance these two is 156.205 N acting exactly opposite to the resultant of these two forces.
If you are uncomfortable with vector sum, you can solve this by resolving the vectors.
place the forces on the cartesian plane in such a way that the 80N force lies in first quadrant and the 100N force lies in the fourth quadrant (each force making 30° with x axis), X axis being the angle bisector of the angle b/w the two forces.
Now you can resolve them into x and y components
for 80N, Fx = 80cos30, Fy = 80sin30
for 100N, Fx = 100cos30, Fy = 100sin30
Clearly you can see the resultant is
Fx = 180cos30
Fy = 20sin30
And magnitude of this force is = ((180cos30)^2 + (20sin30)^2)^1/2 = 156.205
If you need direction, calculate Fy/Fx
