A 165 pound football player assumed a 4 point stance at the line. If his base of support was 30” wide, 38” deep, with the center of gravity 28” off the ground and 16” from the front edge of the base of support, how many inch-pounds would be required to move his center of gravity outside his base of support from a 1) frontal attack, 2) from the side, & 3) from the rear? heelp please ??!!!

(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

false br

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg