1. 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 11.00 N [20.0˚ E of S] is also acting on the mass. The mass of the block is 7.50x10 2 kg. Force Vector Resolution

2. a)List all forces acting on the block. b)List all force directions in standard angle format in your knowns and unknowns. c)Draw a free-body diagram of the forces acting on this block. d)What is the net force acting in the x-direction? e)What is the net force acting in the y-direction? f)What is the resultant (net) force acting on this block? g)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 This is the free-body diagram for the problem we have been given. x F 3

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

6. Force Vector Resolution F 1 = 8.00 N [40.0˚ S of W] = 8.00 N [220.0˚ st.] F 2 = 10.00 N [50.0˚ E of N] = 10.00 N [40.0˚ st.] F 3 = 11.00 N [20.0˚ E of S] = 11.00 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

7. Force Vector Resolution 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. To calculate the resultant force we must use Pythagoras Equation and solve for F R. When adding, we can maintain the same number of decimal places.

8. Force Vector Resolution Since the mass of the object was told to us and is 7.50 x 10 2 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.