Chapter 12, Problem 12.14P

A uniform ladder of length L.and mass m₁ rests against a frictionless wall. The ladder makes an angle 0 with the horizontal, (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when a firelighter of mass m₂, has climbed a distance x along the ladder from the bottom, (b) If the ladder is just on the verge of slipping when the firefighter is a distance d along the ladder from the bottom, what is the coefficient of static friction between ladder and ground?

To determine

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

Answer to Problem 12.14P

The horizontal force exerted by the ground on the ladder $\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$ and the vertical force exerted by the ground on the ladder is $\left(m_2+m_1\right)g$.

Explanation of Solution

Given info: The length of the ladder is $L$, the mass of the ladder is $m_1$, the angle of the ladder with the horizontal is $\theta$, the distance of the fire-fighter from the bottom along the ladder is $x$ and the mass of the fire-fighter is $m_2$.

The free body diagram of the ladder is,

Figure (1)

From figure (1), write the equilibrium condition of forces in the vertical direction.

$N_2-m_{2}g-m_{1}g=0$

Here,

$N_2$ is the normal force applied by the ground on the base of the ladder.

$g$ is the acceleration due the gravity.

Solve the above equation for $N_2$.

$N_2=\left(m_2+m_1\right)g$

Thus, the vertical force exerted by the ground on the base of the ladder is $\left(m_2+m_1\right)g$.

The clockwise direction is taken as positive.

From figure (1), write the equilibrium condition of torque through axis of the ladder about point A.

$\left(m_{2}g\cos \theta \right)\left(x\right)+\left(m_{1}g\cos \theta \right)\left(\frac{L}{2}\right)-N_1\sin \theta \left(L\right)=0$ (1)

Here,

$N_1$ is the normal force applied by the wall on the ladder.

Rearrange the above equation for $N_1$.

$\begin{array}{c}{N}_1=\frac{\left(m_{2}g\cos \theta \right)\left(x\right)+\left(m_{1}g\cos \theta \right)\left(\frac{L}{2}\right)}{L\sin \theta }\\ =g\cot \theta \left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)\\ =\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta \end{array}$

From figure (1), write the equilibrium condition of forces in the horizontal direction.

$F_{\text{f}2}-N_1=0$

Here,

$F_{\text{f}2}$ is the horizontal friction force on the base of the ladder.

Substitute $\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$ for $N_1$ in the above equation.

$\begin{array}{c}{F}_{\text{f}2}-\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta =0\\ F_{\text{f}2}=\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta \end{array}$

Thus, the horizontal frictional force exerted by the ground on the base of the ladder is $\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$.

Conclusion:

Therefore, the horizontal force exerted by the ground on the ladder $\left(\frac{x}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$ and the vertical force exerted by the ground on the ladder is $\left(m_2+m_1\right)g$.

To determine

The coefficient of static friction between the ladder and the ground.

Answer to Problem 12.14P

The coefficient of static friction between the ladder and the ground is $\frac{\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)\cot \theta }{\left(m_2+m_1\right)}$.

Explanation of Solution

Given info: The length of the ladder is $L$, the mass of the ladder is $m_1$, the angle of the ladder with the horizontal is $\theta$, the distance of the fire-fighter from the bottom along the ladder is $d$ and the mass of the fire-fighter is $m_2$.

From equation (1), the equilibrium condition of torque about point A is,

$\left(m_{2}g\cos \theta \right)\left(x\right)+\left(m_{1}g\cos \theta \right)\left(\frac{L}{2}\right)-N_1\sin \theta \left(L\right)=0$

Substitute $d$ for $x$ in the above equation.

$\left(m_{2}g\cos \theta \right)\left(d\right)+\left(m_{1}g\cos \theta \right)\left(\frac{L}{2}\right)-N_1\sin \theta \left(L\right)=0$

Rearrange the above equation for $N_1$.

$\begin{array}{c}{N}_1=\frac{\left(m_{2}g\cos \theta \right)\left(d\right)+\left(m_{1}g\cos \theta \right)\left(\frac{L}{2}\right)}{L\sin \theta }\\ =g\cot \theta \left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)\\ =\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta \end{array}$

From figure (1), write the equilibrium condition of forces in the horizontal direction.

$F_{\text{f}2}-N_1=0$

Substitute $\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$ for $N_1$ in the above equation.

$\begin{array}{c}{F}_{\text{f}1}-\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta =0\\ F_{\text{f}1}=\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta \end{array}$

Thus, the maximum friction between the ground and the ladder is $\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$.

The force of friction between the ladder and ground is,

$F_{\text{f}2}=\mu N_2$

$\mu$ is the coefficient of friction force between the base of the ladder and ground.

Substitute $\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta$ for $F_{\text{f}2}$ and $\left(m_2+m_1\right)g$ for $N_2$ in the above equation.

$\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta =\mu \left(m_2+m_1\right)g$

Rearrange the above equation for $\mu$.

$\begin{array}{c}\mu =\frac{\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)g\cot \theta }{\left(m_2+m_1\right)g}\\ =\frac{\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)\cot \theta }{\left(m_2+m_1\right)}\end{array}$

Conclusion:

Therefore, the coefficient of static friction between the ladder and the ground is $\frac{\left(\frac{d}{L}{m}_2+\frac{m_1}{2}\right)\cot \theta }{\left(m_2+m_1\right)}$.