3 FORCES ACT ON MOVING OBJECT.FIRST FORCE HAS MAGNITUDE OF 80.0N AND DIRECTED DUE NORTH.ANOTHER HAS MAGNITUDE 60.0N AND DIRECTED DUE WEST.WHAT IS THE MAGNITUDE AND DIRECTION OF THIRD FORCE,SUCH THAT THE OBJECT MOVE IN CONSTANT VELOCITY?
To find the magnitude and direction of the third force, we need to consider the forces acting on the object and their resultant force.
Let's break down the given forces:
1. The first force has a magnitude of 80.0N and is directed due north.
2. The second force has a magnitude of 60.0N and is directed due west.
Now, for an object to move at a constant velocity, the net force acting on it must be zero. This means that the resultant force of all the forces acting on the object must be zero.
The forces can be represented as vectors. Since the first force is directed due north and the second force is directed due west, we can represent them as follows:
1. The first force can be represented as a vector A = 80.0N [North].
2. The second force can be represented as a vector B = 60.0N [West].
To find the magnitude and direction of the third force, we need to find the resultant of vectors A and B.
To do this, we can use vector addition. We can add vector A and B to find the resultant vector, which represents the third force:
Resultant vector (R) = A + B
To add these vectors, we can use the cosine law. The magnitude of the resultant vector can be found using the formula:
R^2 = A^2 + B^2 + 2AB * cos(theta)
However, since we want the object to move at a constant velocity, the magnitude of the resultant vector (R) should be zero:
0 = A^2 + B^2 + 2AB * cos(theta)
We can solve this equation to find the angle (theta) at which the third force should act. Once we have the angle, we can use trigonometry to find the magnitude and direction of the third force.
Please provide the magnitude of vector A and vector B, and I will help you calculate the magnitude and direction of the third force.
F2+F1 = -60 + i80.
F2 + F1 + F3 = 0
-60 + i80 + F3 = 0
F3 = 60 - i80
tanAr = -80/60 = -1.333333
Ar = -53.13 = Reference angle.
A = -53.13 + 360 = 306.87 Deg.
F3 = 60 - i80 = 60 / cos306.87 = 100N. @ 306.87 Deg.