Three forces act on a moving object. One force has a magnitude of 61.9 N and is directed due north. Another has a magnitude of 52.0 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.
constant velocity means no acceleration, which means zero net force
so the third force is equal in magnitude to the resultant of the given forces; but opposite in direction
the magnitude of the resultant of the given forces is __ sqrt(61.9^2 + 52.0^2)
the tangent of the angle of the resultant (north of west) is __ 61.9/52.0
To find the magnitude and direction of the third force, we need to use vector addition.
Let's start by representing the given forces graphically. Draw a coordinate system and mark the due north force as 61.9 N in the positive y-direction (upwards) and the due west force as 52.0 N in the negative x-direction (leftwards).
Next, we can calculate the resultant of these two forces using the Pythagorean theorem. The resultant is the vector sum of the given forces.
Using the Pythagorean theorem, we have:
Resultant (R)^2 = (61.9 N)^2 + (52.0 N)^2
R^2 = 3833.61 N^2 + 2704.0 N^2
R^2 = 6537.61 N^2
Taking the square root of both sides, we get:
R = √6537.61 N
R ≈ 80.96 N
So, the magnitude of the third force must be approximately 80.96 N.
Now, let's determine the direction of the third force. We can use trigonometry to find the angle.
Since the third force must be south of east, we can assume it makes an angle θ with the positive x-axis (east) in the fourth quadrant.
To find θ, we can use the tangent function:
tan(θ) = (magnitude of the y-component) / (magnitude of the x-component)
tan(θ) = (61.9 N) / (52.0 N)
θ = tan^(-1)(61.9 N / 52.0 N)
θ ≈ tan^(-1)(1.189)
θ ≈ 50.61° (rounded to two decimal places)
Therefore, the direction of the third force, relative to the positive x-axis (east), is approximately 50.61° south of east.