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?

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