Chapter 8, Problem 107P

(a)

To determine

The maximum distance to the right hand edge of plank that the painter can walk before the tipping of plank.

Answer to Problem 107P

The painter can move up to $0.96\text{m}\text{from the right hand edge}$.

Explanation of Solution

Mass of painter is $61\text{kg}$, length of uniform plank is $6.00\text{m}$, mass of plank is $20.0\text{kg}$, distance to sawhorse from an end of plank is $1.40\text{m}$, mass of paint bucket is $4.0\text{kg}$, and the diameter of bucket is $0.28\text{m}$.

The axis of rotation is passing through the meeting point of plank with right sawhorse.

Write the equation for rotational equilibrium.

$m_{3}g{d}_3-m_{1}g{d}_1-m_{2}g{d}_2=0$

Here, $m_1$ is the mass of painter, $m_2$ is the mass of paint can , $m_3$ is the mass of plank, $g$ is the gravitational acceleration, $d_1$ is the moment arm of weight due to painter, $d_2$ is the moment arm of weight due to can, and $d_3$ is the moment arm of weight due to plank.

Rewrite the above equation in terms of $d_1$.

$d_1=\frac{m_3{d}_3-m_2{d}_2}{m_1}$ (I)

Write the equation to find the distance to painter from the right-hand edge.

$d=1.40\text{m}-d_1$ (II)

Here, $d$ is the distance to painter from the right-hand edge.

Conclusion:

Substitute $20.0\text{kg}$ for $m_3$, $\left(3.00\text{m}-1.40\text{m}\right)$ for $d_3$, $4.0\text{kg}$ for $m_2$, $\left(1.40\text{m}-0.14\text{m}\right)$ for $d_2$, and $61\text{kg}$ for $m_1$ in equation (I) to find $d_1$.

$\begin{array}{c}{d}_1=\frac{\left(\left(20.0\text{kg}\right)\left(3.00\text{m}-1.40\text{m}\right)\right)-\left(\left(4.0\text{kg}\right)\left(1.40\text{m}-0.14\text{m}\right)\right)}{61\text{kg}}\\ =\frac{26.96\text{kg}\cdot \text{m}}{61\text{kg}}\\ =0.44\text{m}\end{array}$

Substitute $0.44\text{m}$ for $d_1$ in equation (II) to find $d$.

$\begin{array}{c}d=1.40\text{m}-0.44\text{m}\\ =0.96\text{m}\end{array}$

Therefore, the painter can move up to $0.96\text{m}\text{from the right hand edge}$.

(b)

To determine

The maximum distance to the left hand edge of plank that the painter can walk before the tipping of plank.

Answer to Problem 107P

The painter can move up to $0.58\text{m}\text{from the left hand edge}$.

Explanation of Solution

Mass of painter is $61\text{kg}$, length of uniform plank is $6.00\text{m}$, mass of plank is $20.0\text{kg}$, distance to sawhorse from an end of plank is $1.40\text{m}$, mass of paint bucket is $4.0\text{kg}$, and the diameter of bucket is $0.28\text{m}$.

The axis of rotation is passing through the meeting point of plank with left sawhorse.

Write the equation for rotational equilibrium.

$m_{1}g{d}_1-m_{2}g{d}_2-m_{3}g{d}_3=0$

Rewrite the above equation in terms of $d_1$.

$d_1=\frac{m_2{d}_2+m_3{d}_3}{m_1}$ (III)

Write the equation to find the distance to painter from the right-hand edge.

$d_{left}=1.40\text{m}-d_1$ (IV)

Here, $d_{left}$ is the distance to painter from the left-hand edge.

Conclusion:

Substitute $20.0\text{kg}$ for $m_3$, $\left(3.00\text{m}-1.40\text{m}\right)$ for $d_3$, $4.0\text{kg}$ for $m_2$, $\left(1.40\text{m}-0.14\text{m}\right)$ for $d_2$, and $61\text{kg}$ for $m_1$ in equation (III) to find $d_1$.

$\begin{array}{c}{d}_1=\frac{\left(\left(4.0\text{kg}\right)\left(1.40\text{m}-0.14\text{m}\right)+\left(\left(20.0\text{kg}\right)\left(3.00\text{m}-1.40\text{m}\right)\right)\right)}{61\text{kg}}\\ =\frac{50.02\text{kg}\cdot \text{m}}{61\text{kg}}\\ =0.82\text{m}\end{array}$

Substitute $0.44\text{m}$ for $d_1$ in equation (IV) to find $d$.

$\begin{array}{c}{d}_{left}=1.40\text{m}-0.82\text{m}\\ =0.58\text{m}\end{array}$

Therefore, the painter can move up to $0.58\text{m}\text{from the left hand edge}$.