Chapter 12, Problem 13P

(a)

To determine

The horizontal and vertical forces the ground exerts on the base of the ladder.

Answer to Problem 13P

The forces in the horizontal direction are, normal force exerts by the wall is $\underset{\_}{268\text{N}}$, the static friction force is $\underset{\_}{268\text{N}}$ and force in the vertical direction is, normal force exerted by the ground on the ladder is $\underset{\_}{1300\text{N}}$.

Explanation of Solution

Write the first condition for equilibrium, the magnitude of the net external force on the object must be equal zero:

  $\sum F_{ext}=0$                                                                                                       (I)

Here, $\sum F_{ext}$ is algebraic sum of all forces acting on the object in the $x$ and $y$ -directions.

Write the second condition for equilibrium, the magnitude of net external torque on the object about a fixpoint must be zero:

  $\sum {\tau }_{ext}=0$                                                                                                      (II)

Here, $\sum {\tau }_{ext}$ is algebraic sum of all torques acting on the object about any point.

The forces acting on the ladder are shown as.

Write the first equilibrium condition for the ladder in $x$ -direction as

  $f_s-n_w=0$                                                                                                   (III)

Here, $f_s$ is the force of static friction, $n_w$ normal force exerts by the wall.

Write the first equilibrium condition for the ladder in $y$ -direction as

  $n_g-m_{1}g-m_{2}g=0$                                                                                     (IV)

Here, $n_g$ is the normal force exerted by the ground on the ladder, $m_1$ is the mass of the ladder, $m_2$ is mass of the firefighter and $g$ is gravitational acceleration.

Write the expression for the weight of the body.

  $F=mg$                                                                                                        (V)

Here, $m$ is the corresponding mass of the body.

If the mass is $m_1$ then the force becomes $F_1$ or if the mass is $m_2$ then the force become $F_2$.

Write the expression for second equilibrium for the ladder in as

  $n_w\left(CF\right)-m_{1}g\left(DO\right)-m_{2}g\left(EO\right)=0$                                                         (VI)

Here, $n_w$ normal force exerts by the wall, $CF$ distance between the line of action of force $n_w$ to the ground, $DO$ is distance between the line of action of force $F_1$ or $m_{1}g$ to the point and $O$, $EO$ distance between line of action of force $F_2$ of $m_{2}g$ to the point $O$.

Substitute $L\sin \theta$ for $CF$, $\frac{L}{2}\cos \theta$ for $DO$ and $x\cos \theta$ for $EO$ in equation (IV).

  $n_w\left(L\sin \theta \right)-m_{1}g\left(\frac{L}{2}\cos \theta \right)-m_{2}g\left(x\cos \theta \right)=0$

Here, $L$ is the length of the ladder and $\theta$ is the angle ladder makes with the horizontal.

Simplify the above expression for $n_w$ as.

  $n_w\left(L\sin \theta \right)=m_{1}g\left(\frac{L}{2}\cos \theta \right)+m_{2}g\left(x\cos \theta \right)$

Re-arrange the terms.

  $n_w=\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x{m}_2\right)}{L\sin \theta }$

Substitute $\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x{m}_2\right)}{L\sin \theta }$ for $n_w$ in equation (I).

  $f_s-\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x{m}_2\right)}{L\sin \theta }=0$

Re-arrange the terms.

  $f_s=\frac{\cos \theta \left(\frac{m_{1}gL}{2}+x{m}_{2}g\right)}{L\sin \theta }$                                                                           (V)

Conclusion:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $60°$ for $\theta$, $500\text{N}$ for $m_{1}g$, $800\text{N}$ for $m_{2}g$, $15\text{m}$ for $L$ and $4\text{m}$ for $x$ in equation (V).

  $\begin{array}{c}{f}_s=\frac{\cos 60°\left(\frac{\left(500\text{N}\right)\left(15\text{m}\right)}{2}+\left(4\text{m}\right)\left(800\text{N}\right)\right)}{\left(15\text{m}\right)\sin 60°}\\ =267.5056\text{N}\\ \approx 268\text{N}\end{array}$

Substitute $268\text{N}$ for $f_s$ in equation (III).

  $\left(268\text{N}\right)-n_w=0$

Simplify the above expression for $n_w$ as.

  $n_w=268\text{N}$

Substitute $500\text{N}$ for $m_{1}g$, $800\text{N}$ for $m_{2}g$, in equation (II)

  $n_g-\left(500\text{N}\right)-\left(800\text{N}\right)=0$

Simplify the above expression for $n_g$ as.

  $n_g=1300\text{N}$

Thus, the forces in horizontal direction are, normal force exerts by the wall is $\underset{\_}{268\text{N}}$,static friction force is $\underset{\_}{268\text{N}}$ and force in vertical direction is, normal force exerted by the ground on the ladder is $\underset{\_}{1300\text{N}}$.

(b)

To determine

The coefficient of static friction at the verge of sliding between ladder and ground.

Answer to Problem 13P

The coefficient of static friction at the verge of sliding between ladder and ground is $\underset{\_}{0.324}$.

Explanation of Solution

At the verge of slipping the force of friction $\left(f_s\right)$ must have its maximum value as.

  ${\left(f_s\right)}_{\max }={\mu }_s{n}_g$                                                                                             (VI)

Here, ${\mu }_s$ is maximum static friction coefficient and ${\left(f_s\right)}_{\max }$ is maximum friction force.

At the verge of slipping, substitute $x^{\prime }$ for $x$ in equation (V).

  ${\left(f_s\right)}_{\max }=\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x^{\prime }{m}_2\right)}{L\sin \theta }$                                                                (VII)

Here, $x^{\prime }$ is the new position of the fire fighter.

Conclusion:

Substitute $\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x^{\prime }{m}_2\right)}{L\sin \theta }$ for ${\left(f_s\right)}_{\max }$ and $g\left(m_1+m_2\right)$ for $n_g$ in equation (VII).

  ${\mu }_s\left(g\left(m_1+m_2\right)\right)=\frac{g\cos \theta \left(\frac{m_{1}L}{2}+x^{\prime }{m}_2\right)}{L\sin \theta }$

Re-arrange the terms,

  ${\mu }_s=\frac{\cos \theta \left(\frac{m_{1}gL}{2}+x^{\prime }{m}_{2}g\right)}{L\left(m_{1}g+m_{2}g\right)\sin \theta }$                                                                     (VIII)

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $60°$ for $\theta$, $500\text{N}$ for $m_{1}g$, $800\text{N}$ for $m_{2}g$, $15.0\text{m}$ for $L$ and $9.00\text{m}$ for $x^{\prime }$ in equation (VIII).

  $\begin{array}{c}{\mu }_s=\frac{\cos 60°\left(\frac{\left(500\text{N}\right)\left(15.0\text{m}\right)}{2}+\left(9.00\text{m}\right)\left(800\text{N}\right)\right)}{\left(\left(500\text{N}\right)+\left(800\text{N}\right)\right)\left(15.0\text{m}\right)\sin 60°}\\ =0.324\end{array}$

Thus, the coefficient of static friction at the verge of sliding between ladder and ground is $\underset{\_}{0.324}$.