Answer:
(1) The front attack is = 1528.56 inch-pounds (2) From the side is = 1237.5 inch- pounds. (3) from the rear is = 1528.32 inch pounds
Explanation:
Solution
Given
The weight of a football player is = 65 pound
Instance = 4 point
Now,
Let us consider that the weight on both hands are the same and weight on both knees are also same.
Let say, weight on each hand be declared as m1, and weight of each leg be m2
Then,
m₁+ m₁ + m₂ +m₂ = 165 pounds
m₁ + m₂ = 82.5 ---------( equation 1)
On the coordinate plane let us assume that the center of gravity G is at the origin (0,0)
so,
2m₁x₁ + 2m₂x₂ /2m₁ + 2m₂ = 0
m₁ * 16 - m₂ * 22 = 0
m₂ = 8 /π m₁------ (2)
Now, let substitute 8 /π m for m₂ in equation 1
m₁ + 8 /π m₁ = 82.5
where m₁ = 47.76 pounds and m₂ = 34.74 pounds
Now,
(a) The front attack
The weight in front that is the one arm is displaced and about the center of gravity
so,
the inch of pound needed is denoted as:
Mfront = 2m₂ * x₂ = 2 * 34.74 *22
The Mfront becomes = 1528.56 inch-pounds
(2) From the side
The weight on one leg and one hand is displaced about the center of gravity G
Mside = 2/2 * y (m₁ +m₂) = 15¹¹ * (82.5)
so,
Mside = 15 * 82.5
Mside = 1237.5 inch- pounds
(c) For the rear attack
Now,
For the real attack, the weight in rear end on the knees is displaced about the center of gravity
Mrear = 2m₁ * x₂ = 2 * 47.76 * 16
Therefore the Mrear = 1528.32 inch pounds
