Three forces act on a moving object. One force has a magnitude of 62.6 N and is directed due north. Another has a magnitude of 45.9 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.
Fn = 62.6
Fw = 45.9
|F| = sqrt (62.6^2+45.9^2) = 77.6
angle north of west = tan^-1 (62.6/45.9)
= 53.8 degrees
so to cancel that
77.6 at 53.8 south of east
To find the magnitude and direction of the third force, we need to determine the net force acting on the object. If the object is moving with a constant velocity, the net force acting on it must be zero.
We can break down the given forces into their north and east components:
Force 1:
Magnitude: 62.6 N
Direction: due north
Force 2:
Magnitude: 45.9 N
Direction: due west
To find the north component of both forces, we use trigonometry:
North component of Force 1 = 62.6 N * cos(0°) = 62.6 N * 1 = 62.6 N
North component of Force 2 = 45.9 N * cos(90°) = 45.9 N * 0 = 0 N
The sum of the north components is 62.6 N + 0 N = 62.6 N
To find the east component of both forces, we again use trigonometry:
East component of Force 1 = 62.6 N * sin(0°) = 62.6 N * 0 = 0 N
East component of Force 2 = 45.9 N * sin(90°) = 45.9 N * 1 = 45.9 N
The sum of the east components is 0 N + 45.9 N = 45.9 N
Since the object is moving with a constant velocity, the net force acting on it must be zero. This means that the sum of the north and east components of the third force must equal zero.
North component of Force 3 + 62.6 N + 0 N = 0 N
North component of Force 3 = -62.6 N
East component of Force 3 + 0 N + 45.9 N = 0 N
East component of Force 3 = -45.9 N
To find the magnitude of the third force, we can use the Pythagorean theorem:
Magnitude of Force 3 = sqrt((-62.6 N)^2 + (-45.9 N)^2)
= sqrt(3912.76 N^2 + 2106.81 N^2)
= sqrt(6019.57 N^2)
= 77.57 N
To find the angle south of east, we can use the inverse tangent function:
Angle = atan((-45.9 N) / (-62.6 N))
= atan(0.735)
≈ 37.1°
Therefore, the magnitude of the third force is approximately 77.57 N, and the direction is approximately 37.1° south of east.
To find the magnitude and direction of the third force, we need to consider the concept of vector addition. The two given forces, one acting due north and the other due west, form a right angle. We can use the Pythagorean theorem to find the resulting magnitude of the two forces combined.
(a) Magnitude of the third force:
Using the Pythagorean theorem, we can find the magnitude of the third force as follows:
Third force magnitude = sqrt[(magnitude of north force)^2 + (magnitude of west force)^2]
= sqrt[(62.6 N)^2 + (45.9 N)^2]
≈ 78.8 N
Therefore, the magnitude of the third force is approximately 78.8 N.
(b) Direction of the third force:
To determine the direction, we can use trigonometry. We will find the angle the third force makes with the positive x-axis (east direction) and express it as a positive angle south of east.
Let's define the angle made by the north force with the positive x-axis as θ. Since the north force is directed due north, it has an angle of 90 degrees (or π/2 radians) with the positive x-axis.
Now, let's find the angle made by the west force with the positive x-axis, which we will call φ. Since the west force is directed due west, it has an angle of 180 degrees (or π radians) with the positive x-axis.
Next, we need to find the resultant angle by adding θ and φ:
Resultant angle = φ - θ
= 180 degrees - 90 degrees
= 90 degrees
Therefore, the resultant angle made by the third force with the positive x-axis is 90 degrees.
Finally, to express the angle as a positive angle south of east, we subtract the resultant angle from 180 degrees:
Angle = 180 degrees - Resultant angle
= 180 degrees - 90 degrees
= 90 degrees
Thus, the direction of the third force, expressed as a positive angle south of east, is 90 degrees.