Chapter 12, Problem 67P

(a)

To determine

To Calculate: The force that exerted by the ladder against the wall.

Answer to Problem 67P

  $F_{\text{By}\text{the}\text{wall}}=0.15\text{kN}$

Explanation of Solution

Given information :

Mass of the ladder $=20.0\text{kg}$

Tension of the wire $=29.4\text{N}$

Breaking point tension of the wire $=200\text{N}$

Mass of the person $80.0\text{kg}$

Formula used:

Torque can be obtained:

  $\tau =rF$

Where, $\tau$ is the torque, F is the applied force and r is the distance between the point where force is applied to the rotational axis.

Calculation:

  

The free-body diagram of the forces acting on the ladder.

Here,

  $l=$ Length of the ladder

  $m$ is the mass of the ladder and $M$ is the mass of the person.

Apply ${\displaystyle \sum \overrightarrow{\tau }}=0$ about an axis through the bottomof the ladder.

  $F_{bywall}\left(l\sin \theta \right)-mg\left(\frac{1}{2}l\cos \theta \right)-Mg\left(\frac{1}{2}l\cos \theta \right)=0$

Solve for $F_{bywall}$ wall and simplify,

  $\begin{array}{c}{F}_{\text{By}\text{the}\text{wall}}=\frac{\left(m+M\right)g\cos \theta }{2\sin \theta }\\ F_{\text{By}\text{the}\text{wall}}=\frac{\left(m+M\right)g}{2\tan \theta }\end{array}$

From the figure given above determine $\theta$ .

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{5.0\text{m}}{1.5\text{m}}\right)\\ \theta ={73.3}^{\circ }\end{array}$

  $\begin{array}{l}{F}_{\text{By}\text{the}\text{wall}}=\frac{\left(20\text{kg}+80\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}{2\tan {73.3}^{\circ }}\\ F_{\text{By}\text{the}\text{wall}}=0.15\text{kN}\end{array}$

Conclusion:

The force that exerted by the ladder against the wall.

  $F_{\text{By}\text{the}\text{wall}}=0.15\text{kN}$

(b)

To determine

To Calculate: The distance thatfrom the bottom end of the ladder that can an $80.0\text{kg}$ person climb.

Answer to Problem 67P

  $L=3.8\text{m}$

Explanation of Solution

Given information :

Mass of the ladder $=20.0\text{kg}$

Tension of the wire $=29.4\text{N}$

Breaking point tension of the wire $=200\text{N}$

Mass of the person $80.0\text{kg}$

Formula used :

From Newton’s second law of motion:

  $\sum F=ma$

Where, m is the mass and a is the acceleration.

Calculation:

Apply ${\displaystyle \sum F_x}=0$ to the ladder,

  $\begin{array}{l}T-F_{\text{By}\text{wall}}=0\\ T=F_{\text{By}\text{wall}}\end{array}$

Apply ${\displaystyle \sum \overrightarrow{\tau }}=0$ about an axis through the bottom of the ladder subject to the condition that $F_{\text{By}\text{wall}}=T_{\text{Max}}$ :

  $T_{Max}\left(l\sin \theta \right)-mg\left(\frac{1}{2}l\cos \theta \right)-Mg\left(L\cos \theta \right)=0$

Here, $L$ is the maximum distance along the ladder that the person can climb without exceeding the maximum tension in the wire.

  $L=\frac{T_{\text{Max}}l\sin \theta -\frac{1}{2}mgl\cos \theta }{Mg\cos \theta }$

  $\begin{array}{l}L=\frac{\left(200\text{N}\right)\left(5.0\text{m}\right)-\frac{1}{2}\left(20\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(1.5\text{m}\right)}{\left(80\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\cos {73.3}^{\circ }}\\ L=3.8\text{m}\end{array}$

Conclusion:

The distance thatfrom the bottom end of the ladder that can an $80.0\text{kg}$ person climb is $L=3.8\text{m}$.