Question

If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground?

Answer 1

To turn while running, the foot must push off the ground, applying a cutting force while instantaneously supporting the weight of the body. This is a vector addition problem where we must account for the normal force on the foot (from the runner's weight) and the acceleration force. Since the weight of the foot is backing up is vertical and the turning force is horizontal, these two force vectors are perpendicular and will form a right triangle.

Only the magnitude matters, so we need not worry about the sign or angle direction.

Now, what's needed is the hypotenuse of a triangle with legs of

 Fg = (65)(10) = 650 N and F = (65)(5) = 325 N.

This calculation can be approximated as √ (3002 + 7002) = 

√ (90000+490000) = √ (580000) = √ (58 x 104) = √ (58) x 102 = 761

Answer 2

Answer:

Force, F = 715.11 newton

Explanation:

Given that,

Mass of the person, m = 65 kg

Acceleration of man during a running turn, $a=5\ m/s^2$

We need to find the magnitude of force experienced by the foot due to the ground. It can be calculated using second law of motion as :

F = ma

$F=65\ kg\times 5\ m/s^2$

F = 325 Newton

Here, the weight of the person and this force force a right angle triangle.

The wight of the person, W = mg

$W=65\ kg\times 9.8\ m/s^2=637\ N$

So, the resultant for is, $F_r=\sqrt{325^2+637^2}$

$F_r=715.11\ N$

So, the f force experienced by the foot due to the ground is 715.11 newtons. Hence, this is the required solution.