An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 3.9 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

Newton's second law states that for every action there is an equal and opposite reaction. Draw a force diagram. It is moving at a constant velocity. Your force diagram for an object moving at a constant velocity is the same as your force diagram for an object at rest, so you can solve the same way. Negative y fOrce of gravity always exists, so the force you are looking for is keeping it up.

I'm not goig to just give you the answer. (ps, I'm a freshman in my 5th week of highschool and I'm in physics, but since we just learned what you're learning, I thought I'd help you.

that did not help at all

F1+F2+F3 = M*a = M*0 = 0.

6.5 - 3.9i + F3 = 0.
F3 = -6.5 + 3.9i = 7.58 N.[31o] N. of W.
= 149o CCW from +x-axis.

To find the direction and magnitude of the third force acting on the object, we can use a vector addition. Since the object is moving with a constant velocity, the net force acting on it must be zero.

Let's break down the given forces into their x and y components:

Force 1: positive x direction (6.5 N)
Force 2: negative y direction (-3.9 N)

Now, let's find the x and y components of the third force:

Let F_net_x be the x-component of the net force, and F_net_y be the y-component of the net force. Since the net force is zero, we have:

F_net_x + Force 1 x-component + 0 = 0
F_net_x = -6.5 N

F_net_y + Force 2 y-component + 0 = 0
F_net_y = 3.9 N

Now we can find the magnitude and direction of the third force using the Pythagorean theorem and trigonometry.

Magnitude of the third force (F_net):

F_net = √(F_net_x^2 + F_net_y^2)
= √((-6.5 N)^2 + (3.9 N)^2)
≈ √(42.25 N^2 + 15.21 N^2)
≈ √(57.46 N^2)
≈ 7.58 N

To determine the direction of the third force, we can use the inverse tangent function (tan^(-1)). Since the x-component is negative and the y-component is positive (from our calculations), the third force will be in the second quadrant of the coordinate system.

Direction of the third force (θ):

θ = tan^(-1)(|F_net_y / F_net_x|)
= tan^(-1)(|3.9 N / -6.5 N|)
≈ tan^(-1)(-0.6)
≈ -31.8°

Therefore, the magnitude of the third force is approximately 7.58 N, and its direction is approximately 31.8° counter-clockwise from the negative x-axis (or in the second quadrant).