2024-1-9 07:45

$\begin{array}{l} ( x_{s} \ +\ x_{k} \ ) \ N_{k} \ =\ mg\ x_{s}\\ N_{k} \ =\ mg\ x_{s} /( x_{s} +x_{k}) \end{array}$

$w = -\mu_k \int N_{k} dx = -mg\mu_k\ x_s \int_{x_{k}^0}^{x_k^1}\frac{dx}{x_s+x} = mg\mu_k \ x_s \ln\frac{x_s+x_k^0}{x_s+x_k^1}$

$x_k^0$， $x_k^1$ 分别为滑动端起始与中止时到杆子中心的距离。

$N_s \mu_s = N_k\mu_k\\ (mg - N_k) \mu_s = N_k \mu_k\\ x_k \mu_s = x_s \mu_k\\ x_k = x_s \mu_k/\mu_s\\$

$x_k= \alpha\ x_s$

$x_s=L, x_k^0=L, x_k^1 = \alpha L\\ w_1 = mg\mu_k L \ln \frac{L+L}{L+ \alpha L}= mg\mu_k L \ln \frac{2}{1+\alpha}$

$x_s= \alpha L, x_k^0=L, x_k^1 = \alpha^2 L\\ w_2 = mg\ \mu_k \alpha L\ln\frac{\alpha L +L}{\alpha L + \alpha^2 L}= mgL\ \mu_k \alpha \ln(1/\alpha)$

$x_s= \alpha^2 L, x_k^0=\alpha L, x_k^1 = \alpha^3 L\\ w_2 = mg\ \mu_k \alpha^2 L\ln\frac{\alpha^2 L +\alpha L}{\alpha^2 L + \alpha^3 L}= mgL\ \mu_k \alpha^2 \ln(1/\alpha)$

$\begin{array}{l} W=\ mg\ \mu _{k} \ L（\ln\frac{2}{1+\alpha } \ +\ \ln( 1/\alpha ) \ \sum _{n=1}^{\infty } \alpha ^{n})\\ =\ mg\ \mu _{k} \ L \left(\ln\frac{2}{1+\alpha } \ +\ \ln( 1/\alpha ) \ \frac{\alpha }{1-\alpha }\right)\\ \end{array}$

$W=mg\ \mu _{k} \ L \left(\ln\frac{2\mu_s}{\mu_s+\mu_k } \ +\ \ \frac{\mu_k}{\mu_s-\mu_k } \ln\frac{\mu_s}{\mu_k} \right)$