## Section16.5Stress and Strain Bootcamp

### Subsection16.5.5Miscellaneous

What is the tallest building one can build with concrete wall of thickness $10\text{ cm}\text{?}$ Data: Density of concrete $= 2,400\text{ kg/m}^3\text{,}$ Breaking stress of concrete compression $= 20\text{ MPa}\text{.}$

Hint

Solution

We need to compare the maximum stress in the wall, which should happen at the bottom of the wall to the breaking stress. Let $w$ denote the thickness of the wall and $h$ the height. Let $\rho$ the density. Let us consider a column of area of cross-section $w\times w$ for calculation purposes. Then, we expect stress of the weight above the bottom part

\begin{equation*} \text{Stress } = \dfrac{m g}{A} = \dfrac{\rho h w^2 g }{ w^2} = \rho h g. \end{equation*}

We see that thickness cancels out from the calculation. Let us equate this to the breaking stress, let us denote that by $B\text{,}$ and solve for $h\text{.}$

\begin{equation*} h = \dfrac{B}{\rho g}. \end{equation*}

Now, we put numerical values in SI units to get

\begin{equation*} h = \dfrac{20\times 10^6}{ 2,400 \times 9.81 } = 850\text{ m}. \end{equation*}

Figure 16.5.29 shows a crane supporting a load of mass $m$ by a cable BC. The boom AB of the crane has mass $M$ and length $L\text{.}$ The boom makes an angle $\theta$ with horizontal diration. The cable BC makes an angle $\phi$ with the boom. (a) Find the tension $T$ in the cable. (b) Find the force at the base of the boom.

Do the problem in symbols first, and then check for the following numbers. $M=1,000\text{ kg}\text{,}$ $L = 20\text{ m}\text{,}$ $m = 4,000\text{ kg}\text{,}$ $\theta = 30^{\circ}\text{,}$ $\phi = 10^{\circ}\text{.}$

Hint 1 (a)

Balance torque on the boom.

Hint 2 (b)

Balance forces on the boom.

$T = 220,000\text{ N}\text{,}$ $A_x = 207,000\text{ N}\text{,}$ $A_y = 124,000\text{ N} \text{.}$

Solution 1 (a)

(a) Note that tension in cable BC is not equal to $mg\text{,}$ the tension in the cable to which load $m$ is attached. We will balance torque on the boom to get the tension in the cable, which we can use in the balance of force equations to obtain the force at the base of the boom.

Figure 16.5.30 helps us find the lever arms we need in the torque calculations. We will use counterclockwise positive.

Balancing the torque on the boom AB gives the following equation.

\begin{equation*} - M g\, \text{(AD)} - m g\, \text{(AE)} + T\, \text{(AF)} = 0, \end{equation*}

where the lever arms are

\begin{gather*} \text{AD} = \dfrac{L}{2}\,\cos\,\theta, \\ \text{AE} = L\,\cos\,\theta, \\ \text{AF} = L\,\sin\,\phi. \end{gather*}

Therefore,

\begin{equation*} T = \dfrac{Mg\cos\,\theta}{2\sin\,\phi} \left( 1 + \dfrac{2m}{M}\right). \end{equation*}

Numerically, with

\begin{align*} \amp M = 1000\text{ kg},\ \ m = 4000\text{ kg}, \\ \amp \theta = 30^\circ,\ \ \phi=10^\circ,\ \ g = 9.81\text{ m/s}^2, \end{align*}

we get

\begin{equation*} T = 220,000\text{ N}. \end{equation*}
Solution 2 (a)

(b) Let $A_x$ and $A_y$ be the components of the force at the base. Balancing of forces on the boom AB gives the following equations.

\begin{align*} \amp A_x - T\,\cos(\theta-\phi) = 0, \\ \amp A_y - Mg - mg - T\,\sin(\theta-\phi)=0, \end{align*}

which give us the components of the force at the base. From these we can find the magnitude and direction of he force at the base.

Numerical values give us

\begin{align*} A_x \amp = 207,000\text{ N}, \\ A_y \amp = 124,000\text{ N}. \end{align*}

A boy is walking on a table of half his mass and getting dangerously close to the edge. If the boy is a distance $x$ from the edge the table will trip about the legs closest to the edge as shown in Figure 16.5.32. You can assume $d = L/5$ and $L=1.5\text{ m}\text{.}$

(a) Find $x$ in terms of the given dimensions in the figure.

(b) What should be the minimum mass of any boy relative to the mass of the table so that the table will not tip at all no matter where the boy stands? That is, if the mass of the table is $M$ and the mass of the boy is $m\text{,}$ give the critical $m/M\text{.}$

Hint

Normal on the far side leg will be zero at the tipping point.

(a) $7.5\text{ cm}\text{,}$ (b) $\lt 1.5\text{.}$

Solution 1 (a)

Figure 16.5.33 shows forces on the table. Just after the tipping occurs, the left leg will no longer be in contact with the flor. Therefore, as the boy moves closer and closer to the right edge, the normal force $N_1$ will decrease, and at the tipping point $N_1=0\text{.}$ Similarly, $F_{s1}=0\text{.}$

The torque of all fiorces about the pivot point O we get

\begin{align*} \amp \tau_{N_2} = 0, \ \ \tau_{F_{s2}} = 0, \\ \amp \tau_{Mg} = Mg \left(\dfrac{L}{2} - d \right), \text{ Counterclockwise}, \\ \amp \tau_{mg} = mg \left(d-x\right), \text{ Clockwise}. \end{align*}

Taking the notation that Counterclockwise is positive and clockwise negative we get the following equation from the net torque on the table.

\begin{equation*} Mg \left(\dfrac{L}{2} - d \right) - mg \left(d-x\right) = 0. \end{equation*}

Solving for $x$ gives

\begin{align*} x \amp = \dfrac{m d - M \left(\dfrac{L}{2} - d \right) }{m} \\ \amp = d - \dfrac{M}{m}\left(\dfrac{L}{2} - d \right). \end{align*}

Using the given data,

\begin{equation*} x = 0.3 - 0.5\left(0.75 - 0.3 \right) = 0.3 - 0.225 = 0.075\text{ m} = 7.5\text{ cm}. \end{equation*}
Solution 2 (b)

(b) The critical condition will correspond to $x=0$ since that will mean we are just at the edge of the table. This gives

\begin{equation*} d - \dfrac{M}{m}\left(\dfrac{L}{2} - d \right) = 0. \end{equation*}

Solving this for $M/m$ we get

\begin{equation*} \dfrac{M}{m} = \dfrac{d}{\left(\dfrac{L}{2} - d \right)}. \end{equation*}

Therefore,

\begin{equation*} \dfrac{m}{M} = \dfrac{L}{2d} - 1 = \dfrac{5}{2}-1 = 1.5. \end{equation*}

This says that if $m/M = 1.5\text{,}$ the boy will tip the table if he sits at the edge, but if his mass is any less, he would not succeed.