Given Information:
Pole weight = w = 26.90 N
Force at point c = Fg
Force at point a = FD
distant between point a and b = 0.750 m
distant between point b and c = 1.31 m
Required Information:
Upward Force at point b = FU = ?
Answer:
Upward Force at point b = FU = 73.88 N
Explanation:
Please refer to the attached diagram,
Fg is the force due to gravity at point c acting downwards,
Fg = w = mg = 26.90 N
Let us take clockwise direction positive and anti-clockwise direction negative then applying the equilibrium condition of torque about point a yields,
ΣM = 0
FU(0.750) - Fg(0.750 + 1.31) = 0
0.750FU - 2.06Fg = 0
0.750FU = 2.06Fg
FU = 2.06Fg/0.750
FU = 2.06(26.90)/0.750
FU = 73.88 N
Bonus:
The other force FD can now be found. Let upward direction of force is positive and downward direction of force is negative.
The the sum of forces acting in the y-axis is given by
ΣFy = 0
-FD + FU - Fg = 0
Where FD is the force at point a acting downwards, FU is the force at point b acting upwards, and Fg is the force due to gravity at point c acting downwards.
-FD + FU - 26.90 = 0
FD = FU - 26.90
FD = 73.88 - 26.90
FD = 46.98 N
