Question

An object acted o by three forces moves with a 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 4.4N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

I went through this problem and solved it and got the answers. 7.85 N for force and 145.9 degrees for direction. Even though i solved it i'm not really grasping the concept behind this question. Can someone please solve this step by step and tell me the purpose of each step?

Answer 1

Organize the information the problem has provided to see how to reach the solution. The key here is that constant velocity indicates 0 net force.

Explanation:

I will start by deconstructing the problem.

An object acted on by three forces moves with a constant velocity

If an object is moving at constant velocity, it means that it has an acceleration of 0. Since `F=ma`, this also means that there is no net force acting on the object.

`therefore F_"net" = 0 N /_ 0^@`

(remember that force not only has a magnitude (in Newtons) but also direction (in degrees). I'm setting the positive x-axis as 0 degrees for this problem, as it's very common.)

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 4.4N and points in the negative y direction.

We are given two forces:
`F_1 = 6.5 N /_ 0^@`
`F_2 = 4.4N /_ 270^@`

Find the direction and magnitude of the third force acting on the object.

We know two of three forces, as well as the net force on the object. We can now calculate the third force.

We should start by adding `F_1` and `F_2` together to see what the current net force on the object is.

`F_1 + F_2`

`magnitude = sqrt(6.5^2 + 4.4^2) = 7.85N`

(note that the two triangles form a right triangle, making this easy. If not, there would be more trigonometry here - calculating the horizontal and vertical magnitudes separately before similarly combining them. Look more into vector addition.)

`angle = tan^-1 (4.4/6.5) = -34.09^@ = 325.91^@`
`F_1 + F_2 = 7.85N angle 325.91^@`

Now that we know `F_1 + F_2`, we can figure out an `F_3` to cancel it out. In our case, we simply need a force of equal magnitude in the opposite direction. The opposite direction of `325.91^@` is `325.91-180 = 145.91^@`.

`therefore F_3 = 7.85N angle 145.91^@`

`square`