Three forces pulling on a body are in equilibrium. The direction of one is due south, the direction of another is 80 degrees east of north, and the third force acts in the direction of 35 degrees west of north. If the magnitude of southerly force is 38 N, find the magnitudes of the other two.
To find the magnitudes of the other two forces, we can use vector addition. The forces are in equilibrium, which means their sum is zero.
Let's break down each force into its horizontal and vertical components. We'll use trigonometry to do this.
1. Southerly force: This force is pulling directly towards the south, so it has no horizontal component and its vertical component is the magnitude itself (38 N).
- Horizontal component: 0 N
- Vertical component: 38 N
2. Force 80 degrees east of north: To determine the components, we need to find the horizontal and vertical lengths of the triangle formed by this force. We'll use the angle of 80 degrees and the magnitude of the force.
- Horizontal component: F_2 * cos(80°)
- Vertical component: F_2 * sin(80°)
3. Force 35 degrees west of north: Like before, we'll find the horizontal and vertical components. This time, we'll use the angle of 35 degrees and the magnitude of the force.
- Horizontal component: F_3 * cos(35°)
- Vertical component: F_3 * sin(35°)
Since the forces are in equilibrium, the sum of the horizontal components and the sum of the vertical components should both equal zero.
Now, let's set up the equations:
Horizontal components: 0 + F_2 * cos(80°) + F_3 * cos(35°) = 0
Vertical components: 38 N + F_2 * sin(80°) - F_3 * sin(35°) = 0
We can simplify these equations to find the magnitudes of F_2 and F_3.
Horizontal components: F_2 * cos(80°) + F_3 * cos(35°) = 0
Vertical components: F_2 * sin(80°) - F_3 * sin(35°) = -38 N
Now we can solve these equations to find the magnitudes of F_2 and F_3.