An athlete of height 1.7 m and mass 50 kg does pushups. Her center of mass is 1.
ID: 1412181 • Letter: A
Question
An athlete of height 1.7 m and mass 50 kg does pushups. Her center of mass is 1.0 m from her feet and her palms are at a distance of 30 cm from her head. She is in an instantaneously static position when her body is i) horizontal (contact only at feet and hands) and ii) when her body her arms are extended vertically by 50 cm. In each case, estimate the forces on each of her hands and feet. In each case assume that she pivots about her ankles, and that her feet are vertically oriented with a distance from floor to ankle of 15 cm.
Explanation / Answer
This is a problem of Newton's second law
Part i)
2Ff+ 2 Fh- W = 0
Torque respect for feet
+ 2 Fh 1.7 – W 1 = 0
Fh = W 1/ 2 1.7
Fh = 50 9.8 / 3.4
Fh = 144.11 N
Calculate from 1
Ff= (W -2 Fh)/2
Ff = ( 50 9.8 – 2 144.11)/2
Ff = 100.89 N
Part ii)
In the latter case the body is tilted
the equation of Newton's second law does not change
2Ff + 2 Fh - W = 0
Torque equation has different distances
ankles pivot axis (feet)
using trigonometry
Sin = (0.50 -0.15)/1.7 = 0.2059
= Sin-1 0.2059
= 11.88
Cos 11.88 = CA /1.7
CA = 1.7 Cos 11.88
CA = 1.6636 m
Cos 11.88 = CA2 /1
CA2 = Cos 11.88
CA2 = 0.9786 m
+2 Fh 1.6636 - W 0.9786 = 0
Fh = W 0.9786 / ( 2 1.6636) = 50 9.8 0.9786/3.3272
Fh = 144.12 N
2Ff + 2 Fh - W = 0
Ff = (W -2Fh)2
Ff = (50 9.8 – 2 144.12)/2
Ff = 100.88 N