Question

q1. in exercise physiology studies, it is sometimes important to determine the location of a person’s center of mass. this determination can be done with the arrangement shown in figure (a). a light plank rests on two scales, which read (f_{g1}) is 380 n and (f_{g2}) is 320 n. a distance of 1.65 m separates the scales. how far from the woman’s feet is her center of mass? q2. suppose the object in figure (b) is the brass base plate of an outdoor sculpture that experiences shear forces in an earthquake. the plate is 0.80 m square and 0.50 cm thick. what is the force exerted on each of its edges if the resulting displacement x is 0.16 mm? (shear modulus of brass is (3.5\times10^{10}) pa)

Explanation:

Step1: Find the total weight of the woman

The total weight $W$ of the woman is the sum of the forces read by the two scales. So $W = F_{g1}+F_{g2}$.
$W=380 + 320=700$ N.

Step2: Set up the torque - equilibrium equation

Let the distance from the woman's feet (where $F_{g2}$ is applied) to the center of mass be $x$. Take the torque about the point where $F_{g2}$ is applied. The torque due to $F_{g1}$ is $F_{g1}\times1.65$ and the torque due to the weight $W$ is $W\times x$. In equilibrium, the sum of torques about this point is zero. So $F_{g1}\times1.65=W\times x$.

Step3: Solve for $x$

Substitute $F_{g1} = 380$ N, $W = 700$ N and $1.65$ m into the equation $F_{g1}\times1.65=W\times x$.
$x=\frac{F_{g1}\times1.65}{W}=\frac{380\times1.65}{700}\approx0.896$ m.

for Q2:

Step1: Calculate the area of the plate

The plate is a square with side length $L = 0.80$ m. The area $A$ of the plate is $A = L^2=(0.80)^2 = 0.64$ m². The thickness $d=0.50$ cm$=0.005$ m and the displacement $x = 0.16$ mm$=0.00016$ m.

Step2: Recall the formula for shear modulus

The shear - modulus formula is $S=\frac{F/A}{x/d}$, where $S$ is the shear modulus, $F$ is the force, $A$ is the area, $x$ is the displacement and $d$ is the thickness. We want to solve for $F$.

Step3: Rearrange the formula for $F$

From $S=\frac{F/A}{x/d}$, we can get $F=\frac{S\times A\times x}{d}$.

Step4: Substitute the values

Substitute $S = 3.5\times10^{10}$ Pa, $A = 0.64$ m², $x = 0.00016$ m and $d = 0.005$ m into the formula.
$F=\frac{3.5\times 10^{10}\times0.64\times0.00016}{0.005}=7.168\times10^{7}$ N.

Answer:

$0.896$ m