Question

a ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. the coefficient of static friction between the ladder and the horizontal surface is (mu_1 = 0.125) and the coefficient of friction between the ladder and the wall is (mu_2 = 0.153). determine the maximum angle (alpha) with the vertical that the ladder can make without falling on the ground.

Explanation:

Step1: Set up force equations

Let the length of the ladder be \(L\), mass be \(m\). The normal force on the ground is \(N_1\) and on the wall is \(N_2\). The frictional force on the ground \(f_1=\mu_1N_1\) and on the wall \(f_2 = \mu_2N_2\).
In the \(x -\)direction: \(N_2=f_1=\mu_1N_1\). In the \(y -\)direction: \(N_1 = mg + f_2=mg+\mu_2N_2\).
Substitute \(N_2=\mu_1N_1\) into \(N_1 = mg+\mu_2N_2\), we get \(N_1=mg+\mu_1\mu_2N_1\), then \(N_1=\frac{mg}{1 - \mu_1\mu_2}\) and \(N_2=\frac{\mu_1mg}{1-\mu_1\mu_2}\).

Step2: Set up torque equation

Take the bottom - of - the - ladder as the pivot point. The torque due to the weight of the ladder \(mg\frac{L}{2}\sin\alpha\), the torque due to \(N_2\) is \(N_2L\cos\alpha\), and the torque due to \(f_2\) is \(f_2L\sin\alpha\).
\(\sum\tau=0\), \(mg\frac{L}{2}\sin\alpha=N_2L\cos\alpha - f_2L\sin\alpha\).
Since \(f_2=\mu_2N_2\) and \(N_2 = \mu_1N_1\), \(mg\frac{L}{2}\sin\alpha=N_2L\cos\alpha-\mu_2N_2L\sin\alpha\).
Substitute \(N_2=\frac{\mu_1mg}{1-\mu_1\mu_2}\) into the torque equation:

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Step3: Calculate the value of \(\alpha\)

Given \(\mu_1 = 0.125\) and \(\mu_2=0.153\), then \(\tan\alpha=\frac{2\times0.125}{1+(0.125\times0.153)}=\frac{0.25}{1 + 0.019125}=\frac{0.25}{1.019125}\approx0.245\).
\(\alpha=\arctan(0.245)\approx13.8^{\circ}\)

Answer:

\(\alpha\approx13.8^{\circ}\)