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 her 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?

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

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 = ?  

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

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