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College Physics by Openstax Chapter 7 Problem 6

How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.33? Assume no friction acts on the wagon.

Figure 7.33 The boy does work on the system of the wagon and the child when he pulls them as shown.

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

In this case, we are given the following values:

\begin{align*}
F & = 50\ \text{N} \\
d & = 30\ \text{m} \\
\theta & = 30^{\circ} 
\end{align*}

Substituting these values into the equation, we have

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 50\ \text{N} \right)\left( 30\ \text{m} \right) \cos 30^{\circ } \\
W & = 1299.0381\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 7 Problem 5

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º with the horizontal. (See Figure 7.32.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.32 A man pushes a crate up a ramp.

Solution:

The Work Done by the Man on the Crate

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

In case where the work done by the man to the crate, the following values are given:

\begin{align*}
F = & \ 500\ \text{N} \\
d = & \ 4\ \text{m} \\
\theta = & \ 0^{\circ} \color{Blue} \left( \text{Force is parallel to displacement} \right) 
\end{align*}

Substituting these values in the equation, we have

\begin{align*}
W = & \ Fd \cos \theta \\
W = & \ \left( 500\ \text{N} \right) \left( 4\ \text{m} \right) \cos 0^{\circ} \\
W = & \ 2000\ \text{N} \cdot \text{m}
\end{align*}

The work done by the man on his body

In this case, the force exerted is counteracted by the weight of the man. This force is directed upward. The displacement is still the 4.0 m along the inclined plane. The angle between the force and the displacement is 70 degrees. 

\begin{align*}
W = & \ Fd \cos \theta \\
W = & \ mg d \cos \theta \\
W = & \ \left( 85.0\ \text{kg} \right) \left( 9.80\ \text{m/s}^2 \right)\left( 4.0\ \text{m} \right) \cos 70^{\circ} \\
W = & \ 1139.6111\ \text{N} \cdot \text{m}
\end{align*}

The Total Work

The total work done by the man is the sum of the work he did on the crate and on his body.

\begin{align*}
W_{T} & = 2000\ \text{N}\cdot \text{m} + 1139.6111\ \text{N}\cdot \text{m} \\
W_{T} & = 3139.6111 \ \text{N}\cdot \text{m} \\
W_{T} & = 3.14 \times 10^{3} \ \text{N}\cdot \text{m} \\
W_{T} & = 3.14 \times 10^{3} \ \text{J}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 7 Problem 4

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?

Solution:

Part A

According to Table 7.1, the energy in 1 gallon of gasoline is 1.2 \times 10^{8}\ \text{J}. Since only 30% of the gasoline goes into useful work, the work done by the friction W_{f} is 

\begin{align*}
W_{f} & =0.30 \left( 2.0\ \text{gal} \right)\left( 1.2 \times 10^{8} \ \text{J/gal}\right) \\
W_{f} & = 72 \times 10^{6}\ \text{J}
\end{align*}

Now, the work done by the friction can also be calculated using the formula below, where F_{f} is the magnitude of the friction force that keeps the car moving at constant speed, and d is the distance traveled by the car.

\begin{align*}
W_{f}=F_{f}d
\end{align*}

We can solve for F_{f} in terms of the other variables.

F_{f} = \frac{W_{f}}{d}

Substituting the given values, we can now solve for the unknown magnitude of the force exerted to keep the car moving at constant speed.

\begin{align*}
F_{f} & = \frac{W_{f}}{d} \\
F_{f} & = \frac{72 \times 10^{6}\ \text{J}}{108\ \text{km}} \\
F_{f} & = \frac{72 \times 10^{6}\ \text{N}\cdot \text{m}}{108 \times 10^{3}\ \text{m}} \\
F_{f} & = 666.6667\ \text{N} \\
F_{f} & = 6.7 \times 10^{2}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

If the required force is directly proportional to speed, then there must be a linear relationship between the required force and speed. In this situation, we can just simply used ratio and proportion to compute for the number of gallons.

\begin{align*}
\frac{2.0\ \text{gal}}{30.0\ \text{m/s}} & = \frac{x}{28.0\ \text{m/s}} \\
x & = \frac{\left( 2.0\ \text{gal} \right)\left( 28.0\ \text{m/s} \right)}{30.0\ \text{m/s}} \\
x & = 1.8667\ \text{gal} \\
x & = 1.9\ \text{gal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 7 Problem 3

(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the elevator car by the gravitational force in this process? (c) What is the total work done on the elevator car?

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

Part A

The force in the cable is equal to the combined effect of the weight of the elevator and the friction that opposes the motion. That is

\begin{align*}
F & = mg + f \\
F & = \left( 1500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)+100\ \text{N} \\
F & = 14800\ \text{N}
\end{align*}

This force in the cable is directed upward. The displacement is also upward, making the angle between the two quantities equal to zero. Thus, \theta = 0.

Substituting these values in the equation, the work done by the cable is

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 14\ 800\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 0^\circ \\
W & = 592\ 000\ \text{J} \\
W & = 5.92 \times 10^{5} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The force due to gravity is equal to the weight of the elevator alone. That is 

\begin{align*}
\text{Weight} & = mg \\
 & = \left( 1\ 500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \\
 & = 14\ 700\ \text{N}
\end{align*}

This force is directed downward, whereas the displacement is directed upward. Therefore, the angle \theta between the two quantities is \theta = 180^\circ.

Substituting these values in the formula for work, we have

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 14\ 700\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 180^\circ \\
W & = -588\ 000\ \text{J} \\
W & = -5.88 \times 10^{5}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

Since the elevator is moving at a constant speed, it is in equilibrium. This means that the net external force experience by the elevator is zero. Therefore, the total work done on the elevator car is 

W_{T} = 0\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer}  \right)

College Physics by Openstax Chapter 7 Problem 2

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)

Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is \Delta PE_{g} = mgh, with h being the increase in height and g the acceleration due to gravity.

W=mgh

We are given the following values: m=75.0\ \text{kg}, g=9.80\ \text{m/s}^2, and h=2.50\ \text{m}.

Substitute the given in the formula.

\begin{align*}
W & = mgh \\
W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\
W & = 1837.5\ \text{Nm} \\
W & = 1837.5\ \text{J} \\
W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is about 1.84 \times 10 ^ {3}\ \text{Joules} .

College Physics by Openstax Chapter 7 Problem 1

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

We are given the following values: F=5.00\ \text{N}, d=0.600\ \text{m}, and \theta=0^\circ.

Substitute the given values in the formula for work.

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 5.00\ \text{N} \right)\left( 0.600\ \text{m} \right) \cos 0^\circ \\
W & = 3.00\ \text{Nm} \\
W & = 3.00\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is 3.00 Joules. Now, we can convert this in unit of kilocalories knowing that 1\ \text{kcal} = 4186\ \text{J}. 

\begin{align*}
W & = 3.00\ \text{J} \\
W & = 3.00\ \text{J}\ \times \ \frac{1\ \text{kcal}}{4186\ \text{J}} \\
W & = 0.000717\ \text{kcal} \\
W & = 7.17 \times 10 ^{-4} \ \text{kcal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done in kilocalories is about 7.17 \times 10 ^{-4}.

College Physics by Openstax Chapter 6 Problem 31

The Speed of the Roller Coaster at the Top of the Loop

Problem:

Modern roller coasters have vertical loops like the one shown in Figure 6.35. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

Figure 6.35 Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than g
 so that the passengers do not lose contact with their seats, nor do they need seat belts to keep them in place.

Solution:

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Homework 4 in MEE 322: Structural Mechanics | Two-Dimensional Stress Analysis

Problem 1

The state of stress at a point is given in the figure. Find τ and τxy directly using force equilibrium. Do NOT use the stress transformation equations.

Problem 2

The state of stress at a point is given in the figure. Find σ and τxy using stress transformation equations.

Problem 3

The state of stress at a point is given in the figure. Find the principal stresses, principal directions and the maximum shear stress using

(a) Eigenvalue problem approach

(b) Stress transformation equations

Problem 4

The shaft shown in the figure has a gear at B with a force of 2098 N in -y and 6456 N in +z applied at its tip. The force along z produces a torque that drives the component attached at C, which produces an equal and opposite torque to that produced at the gear as well as forces on the shaft of 6000 N along +y and +z. The bearing at A can be considered a spherical hinge, whereas the bearing at D can be considered a planar hinge in the y-z plane.

(b) Draw bending moment and torsion diagrams for the shaft and show diagrams of the cross-section of the shaft where the critical points occur, i.e., the locations where the maximum normal stresses due to bending and maximum shear stress due to torsion coincide. Indicate the internal reactions (bending and torsion moments) in this diagram, as well as the locations of the critical points.

(b) If the diameter of the shaft is 33 mm, find the stresses at the critical point and use them to find the maximum shear stress at that location as well as the maximum and minimum principal stresses. Note: the bending normal stress can be taken as σx, while the torsion shear stress can be assumed to be τxy for the effects of this calculation, all other stresses can be assumed to be zero.

(c) It is known that the material of the shaft is such that it will fail if the maximum shear stress reaches 300 MPa. Is the shaft safe? If so, calculate the factor of safety.

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Homework 4 in MEE 322: Structural Mechanics

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Homework 3 in MEE 322 Structural Mechanics | Normal and Shear Stresses Under Combined Loading Part 2

Problem 1

Three forces act on the tip of a L-shaped rod with a cross-sectional radius of 0.5 in.

(a) Determine the normal and shear stress at points A and B and draw the stress cube at those points based on the given coordinate system.

(b) Determine the maximum normal stress on the cross-section and locate the point at which it occurs.

Problem 2

The simply supported solid shaft has a radius of 15 mm and is under static equilibrium. Pulley C has a diameter of 100 mm. The pulleys B and D have the same diameter as each other. The forces on pulley B are at an angle of 45 to the negative z-axis. The forces on pulley C and pulley D are in the z and -y direction. The shaft dimensions are in mm.

(a) Determine the maximum bending and torsional stresses in the shaft.

(b) Locate the point(s) on the cross-section where the bending stress is maximum.

Problem 3

The structural part of a setup to measure net belt tensions in pulleys is shown in the figure. The belt tensions at both sides of the pulley at B (radius 10 cm) are P and F=0.1*P along z and a reaction force is measured from the pulley at C (radius 2 cm), which is connected to a load cell at E with an axial member parallel to x. Pulleys are rigidly attached to rod AD, which is made with a ductile steel rod 60 cm long and 1.27 cm in diameter. Length AB=0.20 m, and length DC=0.15 m. There is a spherical hinge at A and a plane hinge at D. The latter constrains motion in the x-z plane only.

(a) Draw bending moment and torsion diagrams for this structure as functions of the unknown tension P and use them to draw a diagram of the critical section showing internal loads (bending and torsion moments) and the critical points.

(b) Use your results from part a to determine the maximum normal stress due to bending and the maximum shear stress due to torsion in terms of the unknown tension P. Calculate the maximum value that P can have if only bending stresses are considered (with σallow = 350 MPa) and then if only torsion stresses are considered (with τallow = 175 MPa).

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Homework #2 in MEE 322 Structural Mechanics

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