A vaulter holds a 26.90-N pole in equilibrium by exerting an upward force U with her leading hand and a downward force D with he
r trailing hand. Point C is the center of gravity of the pole, a=0.750 m, b=1.31 m, c=2.20 m. What is the magnitudes of U?
2 answers:
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
Answer:
Check attachment for better understanding
Explanation:
Given that
a= 0.75m
b=1.31m
c= 2.2m
Weight of pole is 26.90
Then, Fg = Weight = 26.90
Using Equilibrium of forces
ΣFy = 0
U — D — Fg = 0
U — D = Fg
U — D = 26.9
To calculate U,
We will take moment about point A.
ΣMa = 0
Let the clockwise moment be positive and anti-clockwise be negative
Fg(a+b) — U(a) = 0
26.9(1.31+0.75) —0.75U = 0
26.9(2.06) = 0.75U
0.75U = 55.414
U = 55.414/0.75
U = 73.89 N
To calculate D,
U — D = 26.9
73.89—D =26.9
73.89—26.9 = D
D = 46.99N
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