Three forces are applied to an object, as indicated in the drawing. Force Farrowbold1 has a magnitude of 29.0 newtons (29.0 N) and is directed 30.0° to the left of the +y axis. Force Farrowbold2 has a magnitude of 18.9 N and points along the +x axis. What must be the magnitude and direction (specified by the angle θ in the drawing) of the third force Farrowbold3 such that the vector sum of the three forces is 0 N?
To find the magnitude and direction of the third force (Farrowbold3), such that the vector sum of the three forces is 0 N, we need to use the concept of vector addition.
First, let's break down the given forces into their x and y components. We'll use trigonometry to find these components.
Force Farrowbold1:
Magnitude: 29.0 N
Direction: 30.0° to the left of the +y axis
The x-component (Farrowbold1x) can be found using:
Farrowbold1x = Farrowbold1 * cos(30°)
Farrowbold1x = 29.0 N * cos(30°)
The y-component (Farrowbold1y) can be found using:
Farrowbold1y = Farrowbold1 * sin(30°)
Farrowbold1y = 29.0 N * sin(30°)
Force Farrowbold2:
Magnitude: 18.9 N
Direction: +x axis
Since Farrowbold2 is along the +x axis, its x-component (Farrowbold2x) will be the full magnitude, and its y-component (Farrowbold2y) will be zero.
Now, let's consider the vector sum of the three forces:
0 N = Farrowbold1x + Farrowbold1y + Farrowbold2x + Farrowbold2y + Farrowbold3
Since the vector sum is zero (0 N), the x-components and y-components must cancel each other out.
Equating the x-components, we have:
0 = Farrowbold1x + Farrowbold2x + Farrowbold3x
0 = 29.0 N * cos(30°) + 18.9 N * cos(0°) + Farrowbold3x
Equating the y-components, we have:
0 = Farrowbold1y + Farrowbold2y + Farrowbold3y
0 = 29.0 N * sin(30°) + 0 + Farrowbold3y
To solve for Farrowbold3x and Farrowbold3y, we rearrange the equations:
Farrowbold3x = -29.0 N * cos(30°) - 18.9 N * cos(0°)
Farrowbold3y = -29.0 N * sin(30°)
The magnitude of Farrowbold3 can be found using the Pythagorean theorem:
Magnitude of Farrowbold3 = sqrt((Farrowbold3x)^2 + (Farrowbold3y)^2)
The direction of Farrowbold3 (angle θ) can be found using the inverse tangent (tan^(-1)):
θ = tan^(-1)(Farrowbold3y / Farrowbold3x)
Calculating the values will give you the magnitude and direction of the third force (Farrowbold3) required to cancel out the given forces.
F1 = -14.5i + 25.1j
F2 = 18.9i
Now we want F3 such that
(-14.5i+25.1j)+(18.9i+0j) + F3 = (0,0)
That should help