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 4.4 n and points in the negative y direction. find the direction and magnitude of the third force acting on the object.

magnitude = 7.85 N

direction = 145.9° with respect to the axis x+

Explanation:

Hello

Since the body is moving with constant velocity it means that the sum of all the forces acting on it must be zero

Let F1, F2 and F3 be the three forces:

F1 = 6.5N i

F2 = -4.4 j

F3 = a i + b j

where i { j } stands for the unitary vector pointing towards x+ { y+ }, and a and b are the components of F3 in the x and y directions:

F1 + F2 + F3 = 0

F1 + F2 + F3 = (6.5 N i) + (- 4.4N j) + a i + b j = (6.5 N + a) i  + (b - 4.4 N)j

each component must be zero, therefore:

6.5 + a = 0

-4.4 + b = 0

Therefore, the third force is

F3 = -6.5 N i + 4.4 N j

which has a magnitude of

|F3| = √(6.5^2 + 4.4^2) = √61.61 ≈ 7.85

and the angle (θ) with respect to the x+ axis is :

θ = arcos(-6.5/ 7.85) ≈ ±145.9°

The plus minus sign appeared because cos is an even function. Which sign we must consider? we can imagne it and see that the resulting vector F1+F2 points in the fourth quadrant, therefore F3 must points towards the second quadrant, and the angle 145° is in that quadrant.

Another way is to calculate the angle with respect to the y+ axis, and conclude based on that result the correct angle.

This is important since the component of y could be in the opposite direction and not affect the result of the angle with respect to the x axis

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