Answer
$$\Big(\frac{m_{off}}{m_{on}+m_{off}}\Big)_{max}=0.29$$
Work Step by Step
We can think of the washcloth as having 2 separate parts: one "off" part and one "on" part. We then draw the forces accordingly like the image below.
Consider the "on" part:
- The normal force and the weight of the "on" part are the vertical forces in opposite directions. As there is no vertical movement, $$N=m_{on}g$$
- The weight of the "off" part pulls the "on" part rightward, but it is opposed by static friction $f_s$. As we increase the "off" part, the pulling force $m_{off}g$ increases, till the point it surpasses $f_s^{max}$. So for the washcloth not to slide off,
$$m_{off}^{max}g=f_s^{max}=\mu_s\times N=\mu_s\times m_{on}^{min}g$$ $$m_{off}=\mu_s\times m_{on}$$ $$m_{off}+\mu_sm_{off}=\mu_sm_{on}+\mu_sm_{off}$$ $$m_{off}(\mu_s+1)=\mu_s(m_{on}+m_{off})$$ $$\frac{m_{off}}{m_{on}+m_{off}}=\frac{\mu_s}{\mu_s+1}$$ $$\Big(\frac{m_{off}}{m_{on}+m_{off}}\Big)_{max}=\frac{0.4}{1.4}=0.29$$
