## College Physics (4th Edition)

The fractional length change of the femur when the person moves from standing on two feet to standing on one foot is a decrease of $0.000039$
(a) $Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length We can find the force exerted on the bone by the body weight: $F = mg$ $F = (70~kg)(9.80~m/s^2)$ $F = 686~N$ We can find the vertical compression: $\Delta L = \frac{F~L}{Y~A}$ $\Delta L = \frac{(686~N)(0.50~m)}{(1.1\times 10^{10}~N/m^2)(8.0\times 10^{-4}~m^2)}$ $\Delta L = 3.90\times 10^{-5}~m$ The vertical compression of the bone is $3.90\times 10^{-5}~m$ (b) When the person is standing on both feet, we can assume that each leg supports half the person's weight. Then the force exerted on each bone by the body weight is half of $686~N$. Since the applied force on the bone is reduced by a factor of two, the vertical compression is reduced by a factor of 2. We can find the compression of the bone when standing on two feet: $\Delta L_2 = \frac{1}{2}\times (3.90\times 10^{-5}~m) = 1.95\times 10^{-5}~m$ The length of the bone when standing on two feet: $L_2 = (0.5~m)-(1.95\times 10^{-5}~m) = 0.4999805~m$ The length of the bone when standing on one foot: $L_1 = (0.5~m)-(3.90\times 10^{-5}~m) = 0.499961~m$ We can find the fractional length change when the person moves from standing on two feet to standing on one foot: $\frac{0.499961~m-0.4999805~m}{0.4999805~m} = -0.000039$ The fractional length change of the femur when the person moves from standing on two feet to standing on one foot is a decrease of $0.000039$