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5) a person pulls on a 50 kg desk w/ a 200n force acting at a 30á angle above the horizontal. the desk does not budge.
a) write the equation that describes the forces which act in the x-direction.
b) write the equation that describes the forces which act in the y-direction.
c) determine the x and y components of the force of tension.
d) determine the values of the frictional and normal forces.
6) suppose the person were pushing down at a 30á angle with 200n of force. the desk still does not move.
a) write the equation that describes the forces which act in the x-direction.
b) write the equation that describes the forces which act in the y-direction.
c) determine the values of the frictional and normal forces.

5)
a. The equation that describes the forces which act in the x-direction: 
     Fx = 200 * cos 30 

b. The equation which describes the forces which act in the y-direction: 
     Fy = 200 * sin 30 

c. The x and y components of the force of tension: 
    Tx = Fx = 200 * cos 30  
    Ty = Fy = 200 * sin 30 

d.Since desk does not budge, frictional force = Fx
                                                                        = 200 * cos 30 

                                                 Normal force = 50 * g - Fy
                                                                       = 50 g - 200 * sin 30 

6) Let F_net = 0
a. The equation that describes the forces which act in the x-direction: 
    (200N)cos(30) - F_s = 0

b. The equation that describes the forces which act in the y-direction:
    F_N - (200N)sin(30) - mg = 0

c. The values of friction and normal forces will be:
     Friction force= (200N)cos(30),
     
The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N

Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N

The answer is 225 miles per 4.5 hours. hope this ; )