Chapter 8, Problem 88AP

(a)

To determine

Free body diagram for the ladder.

Answer to Problem 88AP

Explanation of Solution

The following figure shows the free body diagram.

(b)

To determine

The normal force exerted by the bottom of the ladder.

Answer to Problem 88AP

The normal force exerted by the bottom of the ladder is $218\text{N}$.

Explanation of Solution

The sum of forces on the ladder is ${\displaystyle \sum F_y}=0=n_F-w-m_{m}g$ and this expression can be rearranged for the normal force exerted by the bottom of the ladder.

Given info: The weight of the ladder is $1.20\times {10}^2\text{N}$, mass of the monkey is $10.0\text{kg}$, and acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$.

The formula for the normal force exerted by the bottom of the ladder is,

$n_F=w+m_{m}g$

• $w$ is weight of the ladder.
• $m_m$ is mass of the monkey.
• $g$ is acceleration due to gravity.

Substitute $1.20\times {10}^2\text{N}$ for $w$, $10.0\text{kg}$ for $m_m$, and $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find $n_F$.

$\begin{array}{c}{n}_F=1.20\times {10}^2\text{N}+\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\\ =218\text{N}\end{array}$

Thus, the normal force exerted by the bottom of the ladder is $218\text{N}$.

Conclusion:

Therefore, the normal force exerted by the bottom of the ladder is $218\text{N}$.

(c)

To determine

The tension in the rope when the monkey is two-thirds of the way up the ladder.

Answer to Problem 88AP

The tension in the rope when the monkey is two-thirds of the way up the ladder is $72.4\text{N}$.

Explanation of Solution

The sum of torques on the ladder is ${\displaystyle \sum \tau }=0=-w\left(L/2\right)\cos \theta -m_{m}g\left(2L/3\right)\cos \theta +n_{w}L\sin \theta$ and now this expression is rearranged for the normal force on the ladder due to the wall. The forces which act on the ladder along horizontal direction is ${\displaystyle \sum F_x=T-n_w=0}$.

Given info: The weight of the ladder is $1.20\times {10}^2\mathrm{N}$, angle of the ladder with horizontal is $60.0°$, mass of the monkey is $10.0\text{kg}$, and acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$.

The formula for the normal force on the ladder by the wall is,

$n_w=\left(1/2\right)w\cot \theta +\left(2/3\right)m_{m}g\cot \theta$

• $w$ is weight of the ladder.
• $\theta$ is the angle of the ladder with horizontal.
• $m_m$ is mass of the ladder.
• $g$ is acceleration due to gravity.

Substitute $1.20\times {10}^2\text{N}$ for $w$, ${60.0}^{\circ }$ for $\theta$, $10.0\text{kg}$ for $m_m$, and $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find $n_w$.

$\begin{array}{c}{n}_w=\left(1/2\right)\left(1.20\times {10}^2\text{N}\right)\left(\cot 60.0°\right)+\left(2/3\right)\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\left(\cot 60.0°\right)\\ =72.4\text{N}\end{array}$

Thus, the normal force on the ladder due to the surface of the wall is equal to the tension in the rope is $72.4\text{N}$.

Conclusion:

Therefore, the tension in the rope when the monkey is two-thirds of the way up the ladder is $72.4\text{N}$.

(d)

To determine

The maximum distance $d$ that the monkey can climb up the ladder.

Answer to Problem 88AP

The maximum distance $d$ that the monkey can climb up the ladder is $2.41\text{m}$.

Explanation of Solution

The tension in the rope when it ready to break is $T_{\max }=80.0\text{N}$ and the resultant torque on the ladder is $0=-w\left(L/2\right)\cos \theta -m_{m}gx\cos \theta +n_{w}L\sin \theta$, from this expression, the distance of the monkey is obtained by solving the following expression $0=w(L/2)\cos \theta -m_{m}gx\cos \theta +n_{w}L\sin \theta c$.

Given info: The weight of the ladder is $1.20\times {10}^2\mathrm{N}$, angle of the ladder with horizontal is $60.0°$, mass of the monkey is $10.0\text{kg}$, length of the ladder is $3.00\text{m}$, the maximum tension in the rope for  which it can withstand is $80.0\text{N}$, and acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$.

The formula for the maximum distance $d$ that the monkey can climb up the ladder is,

$d=\frac{-w\left(L/2\right)+T_{\max }L\tan \theta }{m_{m}g}$

• $w$ is weight of the ladder.
• $L$ is length of the ladder.
• $T_{\max }$ is maximum tension in the rope.
• $\theta$ is the angle of the ladder with horizontal surface.
• $m_m$ is mass of the monkey.
• $g$ is acceleration due to gravity.

Substitute $1.20\times {10}^2\mathrm{N}$ for $w$, ${60.0}^{\circ }$ for $\theta$, $10.0\text{kg}$ for $m_m$, $80.0\text{N}$ for $T_{\max }$, $3.00\text{m}$ for $L$, and $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find $d$.

$\begin{array}{c}d=\frac{-\left(1.20\times {10}^2\text{N}\right)\left(3.00\text{m}/2\right)+\left(80.0\text{N}\right)\left(3.00\text{m}\right)\left(\tan 60.0°\right)}{\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}\\ =2.41\text{m}\end{array}$

Thus, the maximum distance $d$ that the monkey can climb up the ladder is $2.41\text{m}$.

Conclusion:

Therefore, the maximum distance $d$ that the monkey can climb up the ladder is $2.41\text{m}$.

(e)

To determine

The required alternate analysis if the rope would be removed and the surface be enough rough.

Explanation of Solution

There exists frictional force if the surface would be rough enough and this force acts opposite to the force that being applied on an objects. It is defined as $f={\mu }_{s}n$ where ${\mu }_s$ and $n$ are coefficient of static friction and the normal force.

When the rope tide with lower end of the ladder, the tension in that rope tend to be towards the wall, similarly, when it removed, the frictional force would be directed towards the wall and it do the job of the tension in the rope, the remaining analysis would same as in the previous parts.

For the maximum distance climbed by the monkey is determined by the condition of maximum frictional force it means, the monkey can climb maximum distance if the static friction is large.

Conclusion:

Therefore, to solve this problem, the information of coefficient of static friction is needed.