Question

The figure shows four electrical charges located at the corners of a rectangle. Like charges, you will recall, repel each other while opposite charges attract. Charge B exerts a repulsive force (directly away from B) on charge A of 3.0 N. Charge C exerts an attractive force (directly toward C) on charge A of 6.0 N. Finally, charge D exerts an attractive force of 2.0 N on charge A.

Assuming that forces are vectors, what is the magnitude of the net force F exerted on charge A?

Answer 1

The magnitude of the net force exerted on charge A is 1.0 N.

Step-by-step explanation:

The net force (F) exerted on charge A can be found by summing the individual forces acting on it. Since forces are vectors, we need to consider their magnitudes and directions. The force exerted by charge B is 3.0 N, which is repulsive and directed away from B. The force exerted by charge C is 6.0 N, which is attractive and directed toward C. Finally, the force exerted by charge D is 2.0 N, which is also attractive. To find the net force, we can add up these forces, taking into account their directions: 3.0 N - 6.0 N + 2.0 N = -1.0 N.

Therefore, the magnitude of the net force exerted on charge A is 1.0 N.

Answer 2

Start finding the coordinate of every vector in a rectangular coordinate system with the charge in the origin.

1) Vector Force of B over A, Fba = (-3,0)

2) Vector Force of C over A, Fca = (0. -6)

3) Vector Force of D over A, Fda = (Fda-x. Fda-y)

By similar triangle properties:

[Fda-x] / 2 = 141 / hypotenuse of the triangle

hypotenuse = √(141^2 + 100^2) = 172.86

=> [Fda-x] = 2 * 141 / 172.86 = 1.6314

[Fda-y] / 2 = 100 / 172.86 => [Fda-y] = 2 * 100 / 172.86 = 1.157

=> Fda = (1.6314 , - 1.157)

4) Sum of the vectors:

Net force vector over A: (-3,0) + (0,-6) + (1.6314 , - 1.157) = (-1.3686 , -7.157)

Magnitud = √[ (-1.3686)^2 + (-7.157)^2 ] = 7.287 N

Answer: 7.3 N