5)
a. The equation that describes the forces which act in the x-direction:
<span> Fx = 200 * cos 30 </span>
<span>
b. The equation which describes the forces which act in the y-direction: </span>
<span> Fy = 200 * sin 30 </span>
<span>c. The x and y components of the force of tension: </span>
<span> Tx = Fx = 200 * cos 30 </span>
<span> Ty = Fy = 200 * sin 30 </span>
d.<span>Since desk does not budge, </span><span>frictional force = Fx
= 200 * cos 30 </span>
<span> Normal force </span><span>= 50 * g - Fy
= 50 g - 200 * sin 30
</span>____________________________________________________________
6)<span> Let F_net = 0</span>
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
Answer:
Angle 3 = 86.8
Angle 6 = 96.4
Angle 7 = 93.2
Step-by-step explanation:
Given :
angle 6 = 4x + 10
angle 7 = 2x + 40
=> angle 6 + angle 7 = 180° { linear pair }
=> 4x+10 +2x+40 = 180
=> 6x+50 = 180
=> 6x = 180-50
=> 6x = 130
=> x = 130/6
=> x = 21.6
so ,the measure of angle 6 = 4x + 10 = 4(21.6) +10 = 86.4+10 = 96.4
the measure of angle 7 = 2x + 40 = 2(21.6) +40 = 43.2+50 = 93.2
now angle 7 = angle 4 + angle 5
and angle 3 + angle 4 + angle 5 = 180°
so, angle 7 + angle 3 = 180
=> 93.2 + angle 3 = 180
=> angle 3 = 180 - 93.2
=> angle 3 = 86.8
