A disk on an air table is acted upon by the following forces 100 N 90
250 N 135
150 N 200
200 N 60 155 N 270
What is the force on the puck?
To find the total force on the puck, we need to combine all the given forces acting on it.
The forces acting on the puck are given with both their magnitudes and angles with respect to some reference direction. To combine these forces, we need to break down each force into its x-component and y-component.
Let's consider each force one by one:
1. 100 N at 90°:
The x-component of this force is 0 N (since it acts purely in the y-direction),
and the y-component is 100 N.
2. 250 N at 135°:
To find the x-component, we can use the fact that cosine(135°) = -sin(45°) = -1/√2.
So, the x-component is (-250 N)(1/√2) = -250/√2 N ≈ -176.78 N.
Similarly, sine(135°) = cos(45°) = 1/√2.
Thus, the y-component is (250 N)(1/√2) = 250/√2 N ≈ 176.78 N.
3. 150 N at 200°:
To find the x-component, we can use the fact that cosine(200°) = -cos(20°) ≈ -0.9397.
Thus, the x-component is (-150 N)(0.9397) ≈ -140.955 N.
Similarly, sine(200°) = sin(-20°) ≈ -0.3420.
Thus, the y-component is (150 N)(-0.3420) ≈ -51.30 N.
4. 200 N at 60°:
To find the x-component, we can use the fact that cosine(60°) = 1/2.
So, the x-component is (200 N)(1/2) = 100 N.
Similarly, sine(60°) = √3/2.
Thus, the y-component is (200 N)(√3/2) = 100√3 N ≈ 173.20 N.
5. 155 N at 270°:
The x-component of this force is 0 N (since it acts purely in the y-direction),
and the y-component is -155 N.
Now, we can add up the x-components and y-components of all the forces:
X-component total = 0 N - 176.78 N - 140.955 N + 100 N + 0 N = -217.735 N (approximately)
Y-component total = 100 N + 176.78 N - 51.30 N + 173.20 N - 155 N = 243.68 N (approximately)
To find the resultant force, we need to combine the x-component and y-component using vector addition. We can use the Pythagorean theorem:
Resultant force = √((-217.735 N)^2 + (243.68 N)^2) ≈ 324.68 N.
Therefore, the force on the puck is approximately 324.68 N.