1. Force Vector Resolution
Three forces are acting on a block that is sitting on a frictionless table. A pushing force of 8.00 N [40.0˚ S of W] is acting in one direction while another pushing force of 10.00 N [50.0˚ E of N] is also acting. The third force of N [20.0˚ E of S] is also acting on the mass. The mass of the block is 7.50x102g.

2. Force Vector Resolution
List all forces acting on the block.
List all force directions in standard angle format in your knowns and unknowns.
Draw a free-body diagram of the forces acting on this block.
What is the net force acting in the x-direction?
What is the net force acting in the y-direction?
What is the resultant (net) force acting on this block?
What is the net acceleration of the block, complete with its direction?

3. Force Vector Resolution
This is the free-body diagram for the problem we have been given.

4. Force Vector Resolution
Resolving the x-components
Resolving the y-components
Finding the Resultant
From the tail of the first to the tip of the last vector
From the tail of the first to the tip of the last vector

5. Force Vector Resolution
F1 = 8.00 N [40.0˚ S of W] = 8.00 N [220.0˚ st.]
F2 = 10.00 N [50.0˚ E of N] = 10.00 N [40.0˚ st.]
F3 = N [20.0˚ E of S] = N [290.0˚st.]
List knowns and unknowns
Mathematically resolve the vectors into their x-components and solve for the sum of those components
Mathematically resolve the vectors into their y-components and solve for the sum of those components

6. Force Vector Resolution
To calculate the resultant force we must use Pythagoras Equation and solve for FR. When adding, we can maintain the same number of decimal places.
CAST tells us that this angle must be in quadrants II or IV but the vectors tell is it must be in quadrant IV. Therefore the angle must be 59.93˚S of E.

7. Force Vector Resolution
Since the mass of the object was told to us and is 7.50 x 102 kg, we can calculate the acceleration of the object …
The direction of the acceleration is the same as the direction of the resultant force. Again, therefore the angle must be 59.93˚S of E.