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 the southerly force is 38 N, Find the magnitude of the other two.
To solve this problem, we need to break down each force into its components and then apply the concept of equilibrium.
Let's start by drawing a diagram to visualize the problem.
First, draw a vector representing the southerly force of 38 N pointing directly downwards.
Next, draw a vector representing the force that is 80 degrees east of north. To find the components of this force, we need to calculate the horizontal (x) and vertical (y) components of this force.
The y-component (Fy) can be determined using the sine function:
Fy = F * sin(80)
Fy = 38 N * sin(80°)
Next, the x-component (Fx) can be determined using the cosine function:
Fx = F * cos(80)
Fx = 38 N * cos(80°)
Draw arrows representing the Fx and Fy components of the force. The Fx component should be pointing towards the west, and the Fy component should be pointing towards the north.
Now, draw a vector representing the force that is 35 degrees west of north. Again, we can calculate the horizontal (x) and vertical (y) components of this force.
The y-component (Fy) can be determined using the sine function:
Fy = F * sin(35)
Fy = F * sin(35°)
The x-component (Fx) can be determined using the cosine function:
Fx = F * cos(35)
Fx = F * cos(35°)
Draw arrows representing the Fx and Fy components of the force. The Fx component should be pointing towards the east, and the Fy component should be pointing towards the north-west.
Finally, since the body is in equilibrium, the sum of all the horizontal components and the sum of all the vertical components must be equal to zero.
Equating the horizontal components, we have:
Fx + (-Fx) + (-Fy*sin(80°)) = 0
Simplifying the equation, we get:
2Fx = Fy*sin(80°)
Similarly, equating the vertical components, we have:
(Fy*cos(35°)) + (-38 N) + (-Fy*sin(80°)) = 0
Solving these equations simultaneously will give us the magnitudes of the other two forces.
Substituting the equation for 2Fx into the second equation:
(Fy*cos(35°)) + (-38 N) + (Fy*sin(80°)*cos(35°)) = 0
Now, we can solve this equation to find the value of Fy.
Simplifying the equation and solving for Fy, we get:
Fy = 38 N / (cos(35°) - sin(80°)*cos(35°))
With Fy calculated, we can substitute it back into the equation for 2Fx to find the value of Fx:
2Fx = Fy*sin(80°)
Simplifying the equation and solving for Fx, we get:
Fx = (Fy*sin(80°)) / 2
Therefore, the magnitude of the other two forces can be calculated using these formulas.