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Physics Bootcamp

Samuel J. Ling

Section 16.5 Stress and Strain Bootcamp

Exercises Exercises

Static Equilibrium

1. Angle for a Board to Rest Against a Wall.
Follow the link: Example 16.1.
2. Reading on Two Bathroom Scales For a Man Standing on Plank on Them.
Follow the link: Exercise 16.1.4.1.
7. Forces on a Wooden Plank Resting Against a Support.
Follow the link: Exercise 16.1.4.6.
8. Equilibrium of Supported Bar at an Angle Pushed Down at One End.
Follow the link: Exercise 16.1.4.7.
10. Rolling Angle of a Sphere on an Incline (JEE, 2020).
Follow the link: Exercise 16.1.4.9.

Elasticity and Stress

13. Stretching a Steel Wire - Strain and Stress.
Follow the link: Example 16.23.
16. Stress in an Elevator Cable When Elevator is Accelerating.
Follow the link: Exercise 16.2.5.3.

Types of Stress

24. Force per Unit Area on Container when Water Freezes.
Follow the link: Exercise 16.3.5.5.

Energy in Strained Matrial

25. Strain Energy in Stretched Cable.
Follow the link: Example 16.31.
27. Elastic Energy Stored in a Deformed Jello.
Follow the link: Exercise 16.4.2.

Miscellaneous

28. Tallest Concrete Building.
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.
Answer.
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*}
29. Tension in a Crane Cable.
Figure 16.32 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. \(M=1,000\text{ kg}\text{,}\) \(L = 20\text{ m}\text{,}\) \(m = 4,000\text{ kg}\text{,}\) \(\theta = 30^{\circ}\text{,}\) \(\phi = 10^{\circ}\text{.}\)
Figure 16.32. Figure for Exercise 16.5.29.
Hint 1. (a)
Balance torque on the boom.
Hint 2. (b)
Balance forces on the boom.
Answer.
\(T = 220,000\text{ N}\text{,}\) \(A_x = 207,000\text{ N}\text{,}\) \(A_y = 124,000\text{ N} \text{.}\)
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 16.33 helps us find the lever arms we need in the torque calculations. We will use counterclockwise positive.
Figure 16.33. Figure for Exercise 16.5.29 solution.
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*}
30. Tipping a Table.
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 16.34. 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 \(m/M\text{.}\)
Figure 16.34. Figure for Exercise 16.5.30.
Hint.
Normal on the far side leg will be zero at the tipping point.
Answer.
(a) \(7.5\text{ cm}\text{,}\) (b) \(\lt 1.5\text{.}\)
Solution 1. (a)
Figure 16.35 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, \(F_{s1}=0\text{.}\)
Figure 16.35. Figure for Exercise 16.5.30 solution.
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.
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