Chapter 12, Problem 1P

You are building additional storage space in your garage. You decide to suspend a 10.0-kg sheet of plywood of dimensions 0.600 m wide by 2.25 m long from the ceiling. The plywood will be held in a horizontal orientation by four light vertical chains attached to the plywood at its corners and mounted to the ceiling. After you complete the job of suspending the plywood from the ceiling, you choose three cubic boxes to place on the shelf. Each box is 0.750 m on a side. Box 1 has a mass of 50.0 kg, box 2 has a mass of 100 kg, and box 3 has a mass of 125 kg. The mass of each box is uniformly distributed within the box and each box is centered on the front-to-back width of the shelf. Unbeknownst to you, one of the chains on the right-hand end of your shelf is defective and will break if subjected to a force of more than 700 N. There are six possible arrangements of the three boxes on the shelf, for example, from left to right, Box 1, Box 2, Box 3, and Box 1, Box 3, Box, 2, and four more. Which arrangements are safe (that is, the defective chain will not break if the boxes are arranged in this way), and which arrangements are dangerous?

To determine

The arrangement of the boxes that are safe and the arrangement of the boxes that are dangerous.

Answer to Problem 1P

The arrangement of the boxes that are safe are $2-3-1,3-1-2,3-2-1$ and the arrangement of the boxes that are dangerous are $1-2-3,1-3-2,2-1-3$.

Explanation of Solution

The mass of the sheet of the plywood is $10.0\text{kg}$, the width of the plywood is $0.600\text{m}$, the length of the plywood is $2.25\text{m}$, the side of the cube is $0.750\text{m}$, the mass of the box $1$ is $50.0\text{kg}$, the mass of the box $2$ is $100\text{kg}$, the mass of the box $3$ is $125\text{kg}$ and the force that can be subjected on the chain is $700\text{N}$.

The chains are all hinged at the center only so the angle made by the chain with the plywood is $60°$ and the tension is divided equally on all the four chains irrespective of the position of the boxes.

The free body diagram of the arrangement $1-2-3$ is shown in the figure below.

Figure (1)

Take the moment in the above diagram on the left end.

    $\begin{array}{c}4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_1\left(\frac{x}{2}\right)+m_2\left(x+\frac{x}{2}\right)+m_3\left(2x+\frac{x}{2}\right)\\ 4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_1\left(\frac{x}{2}\right)+m_2\left(\frac{3x}{2}\right)+m_3\left(\frac{5x}{2}\right)\end{array}$

Here, $T$ is the tension in the chain, $M$ is the mass of the plywood, $m_1$ is the mass of box $1$, $m_2$ is the mass  of box $2$, $m_3$ is the mass of the box $3$, $l$ is the length of the plywood and $x$ is the side of the cube.

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$.

$\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=845.3\text{N}\end{array}$

So the tension in each chain in arrangement $1-2-3$ is more than $700\text{N}$ so the chains would break so it is a dangerous arrangement.

The free body diagram of the arrangement $1-3-2$ is shown in the figure below.

Figure (2)

Take the moment in the above diagram on the left end.

    $4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_1\left(\frac{x}{2}\right)+m_3\left(\frac{3x}{2}\right)+m_2\left(\frac{5x}{2}\right)$

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$.

$\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=779.9\text{N}\end{array}$

So the tension in each chain in arrangement $1-3-2$ is more than $700\text{N}$ so the chains would break so it is a dangerous arrangement.

The free body diagram of the arrangement $2-1-3$ is shown in the figure below.

Figure (3)

Take the moment in the above diagram on the left end.

    $4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_2\left(\frac{x}{2}\right)+m_1\left(\frac{3x}{2}\right)+m_3\left(\frac{5x}{2}\right)$

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$.

$\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=739.06\text{N}\end{array}$

So the tension in each chain in arrangement $2-1-3$ is more than $700\text{N}$ so the chains would break so it is a dangerous arrangement.

The free body diagram of the arrangement $2-3-1$ is shown in the figure below.

Figure (4)

Take the moment in the above diagram on the left end.

    $4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_2\left(\frac{x}{2}\right)+m_3\left(\frac{3x}{2}\right)+m_1\left(\frac{5x}{2}\right)$

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$. $\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=693.65\text{N}\end{array}$

So the tension in each chain in arrangement $2-3-1$ is less than $700\text{N}$ so the chains would not break so it is safe arrangement.

The free body diagram of the arrangement $3-1-2$ is shown in the figure below.

Figure (5)

Take the moment in the above diagram on the left end.

    $4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_3\left(\frac{x}{2}\right)+m_1\left(\frac{3x}{2}\right)+m_2\left(\frac{5x}{2}\right)$

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$.

$\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=657.4\text{N}\end{array}$

So the tension in each chain in arrangement $3-1-2$ is less than $700\text{N}$ so the chains would not break so it is safe arrangement.

The free body diagram of the arrangement $3-2-1$ is shown in the figure below.

Figure (6)

Take the moment in the above diagram on the left end.

    $4T\cos \theta \left(l\right)=Mg\left(\frac{l}{2}\right)+m_3\left(\frac{x}{2}\right)+m_2\left(\frac{3x}{2}\right)+m_1\left(\frac{5x}{2}\right)$

Substitute $2.25\text{m}$ for $l$, $60°$ for $\theta$, $10.0\text{kg}$ for $M$, $50.0\text{kg}$ for $m_1$, $100\text{kg}$ for $m_2$, $125\text{kg}$ for $m_3$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.750\text{m}$ for $x$ in above equation to find the value of $T$.

$\begin{array}{c}4T\cos \left(60°\right)\left(2.25\text{m}\right)=\left[\begin{array}{l}\left(10.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{2.25\text{m}}{2}\right)+\left(125\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{\left(0.750\text{m}\right)}{2}\right)\\ +\left(100\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{3\left(0.750\text{m}\right)}{2}\right)\\ +\left(50.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{5\left(0.750\text{m}\right)}{2}\right)\end{array}\right]\\ T=575.75\text{N}\end{array}$

So the tension in each chain in arrangement $3-2-1$ is less than $700\text{N}$ so the chains would not break so it is safe arrangement.

Conclusion:

Therefore, the arrangement of the boxes that are safe are $2-3-1,3-1-2,3-2-1$ and the arrangement of the boxes that are dangerous are $1-2-3,1-3-2,2-1-3$.