Question

SOLVED

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.

Answer 1

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
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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