Final answer:
To algebraically add the given forces of 6 N at 30°, 8 N at 120°, and 10 N at 250°, we must break them into their x and y components, sum those components, and then calculate the magnitude and direction of the resultant force.
Step-by-step explanation:
To answer this question, we must use vector addition to algebraically add the three given forces.
First, we need to break down each force into its horizontal (x-component) and vertical (y-component) components using trigonometric functions:
- Force a: 6 N at 30° from the horizontal
- Force b: 8 N at 120° from the horizontal
- Force c: 10 N at 250° from the horizontal
Each force component can be found using: Fx = Fcos(θ) and Fy = Fsin(θ).
Calculate the components for each force:
- For force a: Fax = 6cos(30°), Fay = 6sin(30°)
- For force b: Fbx = 8cos(120°), Fby = 8sin(120°)
- For force c: Fcx = 10cos(250°), Fcy = 10sin(250°)
After calculating the components, add the x-components together and the y-components together to find the resultant force in the x and y directions:
FRx = Fax + Fbx + Fcx
FRy = Fay + Fby + Fcy
Finally, the magnitude of the resultant force can be calculated using Pythagoras' theorem, and the direction with respect to the horizontal using the tangent function:
|FR| = √(FRx2 + FRy2)
θR = arctan(FRy / FRx)
Now, apply these steps to determine the algebraic sum of the given forces.