Chapter 8, Problem 125P

(a)

To determine

An algebraic solution for $d$ in terms of $M,m,{\mu }_s,L\text{ and }\theta$.

Answer to Problem 125P

An algebraic solution for $d$ in terms of $M,m,{\mu }_s,L\text{ and }\theta$ is $d=\left({\mu }_s\frac{M+m}{M}\tan \theta -\frac{m}{2M}\right)L$.

Explanation of Solution

The system is in equilibrium until the ladder begins to slip.

Write the expression for the Newton’s second law for the equilibrium.

$\Sigma F_x=0$ (I)

Here, $\Sigma F_x$ is the net external force acting on the system in the $x$ direction.

Write the expression for $\Sigma F_x$.

$\Sigma F_x=N_w-f$

Here, $N_w$ is the normal force acting on the ladder due to the wall and $f$ is the frictional force acting on the base of the ladder due to the floor

Put the above equation in equation (I).

$\begin{array}{c}{N}_w-f=0\\ N_w=f\end{array}$ (II)

The frictional force has its maximum possible magnitude at the person’s highest point.

Write the equation for the maximum possible magnitude of frictional force.

$f={\mu }_s{N}_f$

Here, ${\mu }_s$ is the coefficient of static friction between the floor and the ladder and $N_f$ is the normal force on the ladder due to the floor

Put the above equation in equation (II).

$N_w={\mu }_s{N}_f$ (III)

The net external force on the system in the $y$ direction must be zero.

$\Sigma F_y=0$ (IV)

Here, $\Sigma F_y$ is the net external force on the system in the $y$ direction

Write the expression for $\Sigma F_y$.

$\Sigma F_y=N_f-Mg-mg$

Here, $M$ is the mass of the person, $g$ is the acceleration due to gravity and $m$ is the mass of the ladder

Put the above equation in equation (IV) and rewrite it for $N_f$.

$\begin{array}{c}{N}_f-Mg-mg=0\\ N_f=Mg+mg\\ =\left(M+m\right)g\end{array}$

Put the above equation in equation (III).

$N_w={\mu }_s\left(M+m\right)g$ (V)

Choose the axis of rotation at the contact point between the ladder and the floor.

The net torque about the rotation axis is zero in equilibrium.

$\Sigma \tau =0$ (VI)

Here, $\Sigma \tau$ is the net external torque on the system

Write the expression for $\Sigma \tau$.

$\begin{array}{c}\Sigma \tau =N_f\left(0\right)+f\left(0\right)-N_{w}L\sin \theta +mg\left(\frac{1}{2}L\cos \theta \right)+Mgd\cos \theta \\ =-N_{w}L\sin \theta +mg\left(\frac{1}{2}L\cos \theta \right)+Mgd\cos \theta \end{array}$

Here, $L$ is the length of the ladder, $\theta$ is the angle between the ladder and the floor and $d$ is the distance from the bottom of the ladder to the point where the person stands

Put the above equation in equation (VI) and rewrite it for $d$.

$\begin{array}{c}-N_{w}L\sin \theta +mg\left(\frac{1}{2}L\cos \theta \right)+Mgd\cos \theta =0\\ d=\frac{N_{w}L\sin \theta -\frac{1}{2}mgL\cos \theta }{Mg\cos \theta }\end{array}$

Put equation (V) in the above equation.

$\begin{array}{c}d=\frac{{\mu }_s\left(M+m\right)gL\sin \theta -\frac{1}{2}mgL\cos \theta }{Mg\cos \theta }\\ =\frac{\left[{\mu }_s\left(M+m\right)\sin \theta -\frac{m}{2}\cos \theta \right]L}{M\cos \theta }\\ =\frac{{\mu }_s\left(M+m\right)\sin \theta }{M\cos \theta }+\frac{\frac{m}{2}\cos \theta }{M\cos \theta }\\ =\left({\mu }_s\frac{M+m}{M}\tan \theta -\frac{m}{2M}\right)L\end{array}$ (VII)

Conclusion:

Therefore, the algebraic solution for $d$ in terms of $M,m,{\mu }_s,L\text{ and }\theta$ is $d=\left({\mu }_s\frac{M+m}{M}\tan \theta -\frac{m}{2M}\right)L$.

(b)

To determine

Use the expression derived in part (a) to show that placing the ladder at a larger angle $\theta$ enables the person to climb farther up the ladder without having it slip, all other things being equal.

Answer to Problem 125P

It is shown using the expression derived in part (a) that placing the ladder at a larger angle $\theta$ enables the person to climb farther up the ladder without having it slip, all other things being equal.

Explanation of Solution

Refer to equation (VII) derived in part (a).

The value of $d$ increases with increase in $\tan \theta$ according to equation (VII). The value of $\tan \theta$ increases on the interval $0\le \theta \le 90°$. This implies placing the ladder at a larger angle (that is more nearly vertical) allows the person to climb farther up the ladder without having it slip.

Conclusion:

Thus, it is shown using the expression derived in part (a) that placing the ladder at a larger angle $\theta$ enables the person to climb farther up the ladder without having it slip, all other things being equal.

(c)

To determine

The maximum angle that enables the person to climb all the way to the top of the ladder.

Answer to Problem 125P

The maximum angle that enables the person to climb all the way to the top of the ladder is $63°$.

Explanation of Solution

To climb up to the top of the ladder, set the value of $d$ to the length of the ladder.

$d=L$

Put the above equation in equation (VII).

$\begin{array}{c}L=\left({\mu }_s\frac{M+m}{M}\tan \theta -\frac{m}{2M}\right)L\\ L+L\left(\frac{m}{2M}\right)=\left({\mu }_s\frac{M+m}{M}\tan \theta \right)L\\ 1+\frac{m}{2M}={\mu }_s\frac{M+m}{M}\tan \theta \\ \frac{2M+m}{2}={\mu }_s\left(M+m\right)\tan \theta \\ \tan \theta =\frac{2M+m}{2{\mu }_s\left(M+m\right)}\end{array}$

Rewrite the above equation for $\theta$.

$\theta ={\tan }^{-1}\frac{2M+m}{2{\mu }_s\left(M+m\right)}$ (VIII)

Conclusion:

Given that the value of ${\mu }_s$ is $0.45$ , mass of the person is $60.0\text{ kg}$ and the  mass of the ladder is $15.0\text{ kg}$.

Substitute $60.0\text{ kg}$ for $M$ , $15.0\text{ kg}$ for $m$ and $0.45$ for ${\mu }_s$ in equation (VIII) to find the value of $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\frac{2\left(60.0\text{ kg}\right)+15.0\text{ kg}}{2\left(0.45\right)\left(60.0\text{ kg}+15.0\text{ kg}\right)}\\ =63°\end{array}$

Therefore, the maximum angle that enables the person to climb all the way to the top of the ladder is $63°$.