Five forces act on an object.
(1) 63 N at 90°
(2) 40 N at 0°
(3) 75 N at 270°
(4) 40 N at 180°
(5) 50 N at 60°
What are the magnitude and direction of a sixth force that would produce equilibrium?
To find the magnitude and direction of the sixth force that would produce equilibrium, we need to first determine the sum of all the forces acting on the object. For equilibrium, the net force acting on the object must be zero.
To calculate the net force, we need to calculate the horizontal and vertical components of each force and then sum them separately.
Let's start by finding the horizontal and vertical components for each force:
Force (1):
- Horizontal component: 63 N * cos(90°) = 0 N
- Vertical component: 63 N * sin(90°) = 63 N
Force (2):
- Horizontal component: 40 N * cos(0°) = 40 N
- Vertical component: 40 N * sin(0°) = 0 N
Force (3):
- Horizontal component: 75 N * cos(270°) = 0 N
- Vertical component: 75 N * sin(270°) = -75 N
Force (4):
- Horizontal component: 40 N * cos(180°) = -40 N
- Vertical component: 40 N * sin(180°) = 0 N
Force (5):
- Horizontal component: 50 N * cos(60°) = 25 N
- Vertical component: 50 N * sin(60°) = 43.3 N
Now, let's sum the horizontal and vertical components separately:
Horizontal component sum: 0 N + 40 N + 0 N - 40 N + 25 N = 25 N
Vertical component sum: 63 N + 0 N - 75 N + 0 N + 43.3 N = 31.3 N
To achieve equilibrium, the horizontal and vertical components of the net force should be zero. Since the horizontal component sum is 25 N and the vertical component sum is 31.3 N, we need to add a sixth force to balance these values.
To find the magnitude and direction of the sixth force, we can use the Pythagorean theorem and trigonometry:
Magnitude of the sixth force: √(25 N)^2 + (31.3 N)^2 ≈ 39.7 N
Direction of the sixth force:
- We can use the inverse tangent function to find the direction (angle) of the sixth force:
- Angle = atan(vertical component sum / horizontal component sum)
- Angle ≈ atan(31.3 N / 25 N) ≈ 51.1°
Therefore, the magnitude of the sixth force that would produce equilibrium is approximately 39.7 N, and its direction is approximately 51.1° relative to the horizontal axis.