Question

practice it
use the worked example above to help you solve this problem. in this example, we show how to find the
location of a persons center of gravity. suppose your lab partner has a height ( l ) of 185 cm (6 ft, 1 in) and
a weight ( w ) of 789 n (177 lb). you can determine the position of his center of gravity by having him stretch
out on a uniform board supported at one end by a scale, as shown in the figure. if the boards weight ( w _ { b } ) is 52 n and the scale reading ( f ) is ( 4.58 \times 10 ^ { 2 } mathrm { n } ), find the distance of your lab partners center of gravity
from the left end of the board.
enter a number. differs from the correct answer by more than 10%. double check your calculations. cm

Explanation:

Step1: Set up the torque equation

Take the left - hand end of the board as the pivot. The torque due to the scale force \(F\) is \(F\times L\) (counter - clockwise torque), the torque due to the board's weight \(w_b\) is \(w_b\times\frac{L}{2}\) (clockwise torque), and the torque due to the partner's weight \(w\) is \(w\times x\) (clockwise torque). According to the torque equilibrium \(\sum\tau = 0\), we have \(F\times L=w_b\times\frac{L}{2}+w\times x\).

Step2: Solve for \(x\)

First, express the formula for \(x\):

$$
LATEXBLOCK0
$$

Given \(L = 185\mathrm{cm}\), \(w_b=52\mathrm{N}\), \(w = 789\mathrm{N}\), \(F = 458\mathrm{N}\)

$$
LATEXBLOCK1
$$

$$x=\frac{79920}{789}\approx101\mathrm{cm}$$

Answer:

\(101\mathrm{cm}\)