The staircase rests on a vertical wall and floor. At what values of the angle between the ladder and the floor can it stand if the coefficients of friction of the ladder on the floor and on the wall are equal to $\mu_1$ and $\mu_2$, respectively?
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" alt="2.8.34" width="300" />
<figcaption>
For problem 2.8.34
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
See the following figure...
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="2.8.34" width="350" />
<figcaption>
Force analysis
</figcaption>
</figure>
</center>
@@ -75,22 +75,22 @@
<p>
Let's consider the Newton Second Law for both axis. Firstly, for $x$-axis,
−
$$N_1+f_r_2 = mg$$
+
$$N_1+f_{r2} = mg$$
$$N_1+\mu_2 N_2 = mg \;(1)$$
For $y$-axis,
−
$$N_2 = f_r_1$$
+
$$N_2 = f_{r1}$$
$$N_2 = \mu_1 N_1 \;(2)$$
Substituting $(2)$ into $(1)$:
$$N_1 (1+\mu_1 \mu_2) = mg \;(3)$$
Applying Newton Second Law for the rotation about the pivot point,
$$\sum\vec{\tau} = \vec{0}$$
−
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_r_1 l \sin{\pi-\alpha} = 0$$
+
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_{r1} l \sin{\pi-\alpha} = 0$$
where $l$ is the staircase's length. Now, for avoiding that staircase slides over ground $\sum\tau\leq 0$, then
−
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_r_1 l \sin{\pi-\alpha}\leq 0$$
+
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_{r1} l \sin{\pi-\alpha}\leq 0$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-\mu_1 N_1 l \sin{\pi-\alpha}\leq 0$$
Developing the above expression,
$$mg \geq 2N_1(1-\mu_1 \tan{\alpha}) \;(4)$$
Putting $(3)$ into $(4)$, and separating $\tan{\alpha}$ term,
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
The staircase rests on a vertical wall and floor. At what values of the angle between the ladder and the floor can it stand if the coefficients of friction of the ladder on the floor and on the wall are equal to $\mu_1$ and $\mu_2$, respectively?
The staircase rests on a vertical wall and floor. At what values of the angle between the ladder and the floor can it stand if the coefficients of friction of the ladder on the floor and on the wall are equal to $\mu_1$ and $\mu_2$, respectively?
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" alt="2.8.34" width="300" />
loading="lazy" alt="2.8.34" width="300" />
<figcaption>
<figcaption>
For problem 2.8.34
For problem 2.8.34
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
See the following figure...
See the following figure...
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="draw.png"
<img src="draw.png"
loading="lazy" alt="2.8.34" width="350" />
loading="lazy" alt="2.8.34" width="350" />
<figcaption>
<figcaption>
Force analysis
Force analysis
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
@@ -75,22 +75,22 @@
<p>
<p>
Let's consider the Newton Second Law for both axis. Firstly, for $x$-axis,
Let's consider the Newton Second Law for both axis. Firstly, for $x$-axis,
$$N_1+f_r_2 = mg$$
$$N_1+f_{r2} = mg$$
$$N_1+\mu_2 N_2 = mg \;(1)$$
$$N_1+\mu_2 N_2 = mg \;(1)$$
For $y$-axis,
For $y$-axis,
$$N_2 = f_r_1$$
$$N_2 = f_{r1}$$
$$N_2 = \mu_1 N_1 \;(2)$$
$$N_2 = \mu_1 N_1 \;(2)$$
Substituting $(2)$ into $(1)$:
Substituting $(2)$ into $(1)$:
$$N_1 (1+\mu_1 \mu_2) = mg \;(3)$$
$$N_1 (1+\mu_1 \mu_2) = mg \;(3)$$
Applying Newton Second Law for the rotation about the pivot point,
Applying Newton Second Law for the rotation about the pivot point,
$$\sum\vec{\tau} = \vec{0}$$
$$\sum\vec{\tau} = \vec{0}$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_r_1 l \sin{\pi-\alpha} = 0$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_{r1} l \sin{\pi-\alpha} = 0$$
where $l$ is the staircase's length. Now, for avoiding that staircase slides over ground $\sum\tau\leq 0$, then
where $l$ is the staircase's length. Now, for avoiding that staircase slides over ground $\sum\tau\leq 0$, then
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_r_1 l \sin{\pi-\alpha}\leq 0$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-f_{r1} l \sin{\pi-\alpha}\leq 0$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-\mu_1 N_1 l \sin{\pi-\alpha}\leq 0$$
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-\mu_1 N_1 l \sin{\pi-\alpha}\leq 0$$
Developing the above expression,
Developing the above expression,
$$mg \geq 2N_1(1-\mu_1 \tan{\alpha}) \;(4)$$
$$mg \geq 2N_1(1-\mu_1 \tan{\alpha}) \;(4)$$
Putting $(3)$ into $(4)$, and separating $\tan{\alpha}$ term,
Putting $(3)$ into $(4)$, and separating $\tan{\alpha}$ term,
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
