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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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Homework 2 in MEE 322: Structural Mechanics | Normal and Shear Stress Under Combined Loading


Problem 1

The shaft with a circular cross-section is supported by two bearings at O and C. The bearings do not exert any moments or axial force on the shaft, and they act to constrain motion along the x and y axes. Find the minimum diameter required for the shaft if the maximum normal stress in the shaft cannot exceed 250~ \text{MPa}. All dimensions are in \text{mm}.

Problem 2

The beam ABCD shown in the figure is simply supported at A and D, and has a circular cross-section with a diameter of 80~\text{mm}. The distributed force acts at an angle of 60^\circ to the Z axis and has components only along the Z and X axis. The 1.2~\text{kN} force acts parallel to the Z axis and the 1.5~\text{kN} force acts parallel the X axis. AB = 0.6~ \text{m}, BC = 0.4~ \text{m} and CD = 1~ \text{m}.

(a) Draw shear force and bending moment diagrams for the XY and YZ planes.

(b) Determine the maximum tensile and compressive bending stress in the beam and show their locations on the cross-section of the beam.

Problem 3

The following steel structure will be made with a round bar 35~\text{mm} in diameter, such that section A-C is parallel to the y-axis and section B-E is parallel to x-axis.

An unknown moment T parallel to y is applied at point A, where there is also a spherical (ball) hinge, which constraints translations along x, y, and z axes. The plane hinge at C is contained in the x-z plane, which means that it constrains translations along the x and z axes. The force of 1000~\text{N} at D goes in -z, the force of 500~\text{N} at D goes in -y and the force of 500~\text{N} at E goes in -x. Given these conditions and the coordinate system provided, find:

(a) Diagrams of axial force, bending moment, and torsion moments for sections A-C and B-E. Use the given coordinate system to label the planes where you are making your internal reaction diagrams (x-y, x-z or z-y) and draw the corresponding axes.

(b) Calculate the maximum normal stress due to bending in this structure. Show in a diagram the point(s) in the cross-section of the structure where this stress occurs, and include the bending moments in each axis, the resultant moment, and the neutral axis. Use the given coordinate system to show the orientation of your diagram.

(c) Calculate the maximum shear stress due to torsion in this structure. Show in a diagram the point(s) in the cross-section of the structure where this stress occurs, including the torsion moment. Use the given coordinate system to show the orientation of your diagram.


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Homework 2 in MEE 322: Structural Mechanics

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College Physics by Openstax Chapter 6 Problem 30

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads


Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?


Solution:

Part A

The formula for an ideally banked curve is

\tan \theta = \frac{v^2}{rg} 

Solving for v in terms of all the other variables, we have

v= \sqrt{rg \tan \theta}

For this problem, we are given

  • radius, r=100\ \text{m}
  • acceleration due to gravity, g=9.81\ \text{m/s}^2
  • banking angle, \theta = 15.0^\circ

Substituting all these values in the formula, we have

\begin{align*}
v & = \sqrt{rg \tan \theta} \\ 
v & = \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)\tan 15.0^\circ }\\
v & = 16.2129\ \text{m/s} \\
v & = 16.2\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Let us draw the free-body diagram of the car.

Summing forces in the vertical direction, we have

N \cos \theta + f \sin \theta -w= 0  \quad  \quad \quad \color{Blue} \text{Equation 1}

Summing forces in the horizontal directions taking to the left as the positive since the centripetal force is directed this way, we have

N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2}

We are given the following quantities:

  • radius of curvature, r=100\ \text{meters}
  • banking angle, \theta = 15.0^\circ
  • velocity, v=20\ \text{km/h} = 5.5556\ \text{m/s}
  • We also know that the friction, f=\mu_{s} N and weight, w=mg

We now use equation 1 to solve for N in terms of the other variables.

\begin{align*}
N \cos \theta + f \sin \theta -w & = 0 \\
N \cos \theta +\mu_s N\sin \theta  & = mg \\
N \left( \cos \theta + \mu_s \sin\theta \right) & =mg \\
N & = \frac{mg}{\cos \theta + \mu_s \sin\theta} \ \qquad \ \color{Blue} \left( \text{Equation 3} \right)
\end{align*}

We also solve for N in equation 2.

\begin{align*}
N \sin \theta - \mu_s N \cos \theta & = m \frac{v^2}{r} \\
N \left( \sin \theta-\mu_s \cos \theta \right) & = m \frac{v^2}{r} \\
N & = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \ \qquad \ \color{Blue} \left( \text{Equation 4} \right)
\end{align*}

Now, we have two equations of N. We now equate these two equations.

\frac{mg}{\cos \theta + \mu_s \sin\theta}  = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)}

We can now use this equation to solve for \mu_s.

\begin{align*}
\frac{\bcancel{m}g}{\cos \theta + \mu_s \sin\theta}  & = \frac{\bcancel{m}v^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \\
rg\left( \sin \theta-\mu_s \cos \theta \right) & = v^2 \left( \cos \theta + \mu_s \sin\theta \right) \\
rg \sin \theta - \mu_s rg \cos \theta & = v^2 \cos \theta +\mu_s v^2 \sin \theta \\
\mu_s v^2 \sin \theta + \mu _s rg \cos \theta & = rg \sin \theta - v^2 \cos \theta \\
\mu _s \left( v^2 \sin \theta + rg \cos \theta \right) & = rg \sin \theta - v^2 \cos \theta \\
\mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta}
\end{align*}

Now that we have an equation for \mu_s, we can substitute the given values.

\begin{align*}
\mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \\
\mu_s & = \frac{100\ \text{m}(9.81\ \text{m/s}^2) \sin 15.0^\circ -\left( 5.5556\ \text{m/s} \right)^2 \cos 15.0^\circ }{\left( 5.5556\ \text{m/s} \right)^2 \sin 15.0^\circ +100\ \text{m}\left( 9.81\ \text{m/s}^2 \right) \cos 15.0^\circ } \\
\mu_s & = 0.2345 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts


Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle \theta below the horizontal will the cage hang when the centripetal acceleration is  10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle 10g should be.)

Figure 6.34 (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, a_c, is calculated using the formula a_c = r \omega ^2. Solving for the angular velocity, \omega, in terms of the other variables, we should come up with

\omega = \sqrt{\frac{a_c}{r}}

We are given the following values:

  • centripetal acceleration, a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2
  • radius of curvature, r = 15.0\ \text{m}

Substituting the given values into the equation,

\begin{align*}
\omega & = \sqrt{\frac{a_c}{r}} \\ \\
\omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\
\omega & = 2.5573\ \text{rad/sec} \\ \\
\omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29
The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

\begin{align*}
\sum_{}^{} F_y & = 0 \\ \\
F_{arm} \sin \theta-w & = 0 \\ \\
F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1}
\end{align*}

Now, summing forces in the horizontal direction, taking into account that F_c is the centripetal force which is the net force. That is,

\begin{align*}
F_c & = m a_c
\end{align*}

We know that F_c is equal to the horizontal component of the force F_{arm}. That is F_c = F_{arm} \cos \theta. Therefore,

\begin{align*}
F_{arm} \cos \theta & = m a_c \\
\end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at 10g. Also, we note that the weight w is equal to mg. So, we have

\begin{align*}
F_{arm} \cos \theta & = m a_c \\ \\
\frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\
\frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\
\end{align*}



From here, we are going to use the trigonometric identity \displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We can also cancel m, and g since they can be found on both sides of the equation.

\begin{align*}
\frac{1}{\tan \theta} & = 10 \\ \\
\tan \theta & = \frac{1}{10} \\ \\
\theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\
\theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Homework 1 in MEE 322: Structural Mechanics | Internal Reactions in 3D and 2D Bending


Problem 1

The structure shown in Fig. 1.1 is made with a steel bar that has a rectangular cross-section 5 cm tall and 10 cm wide.

Fig 1.1. Frame supporting point and distributed load

Given the geometry, loads and supports in the structure draw the bending moment diagrams for segments A-C and D-E of the structure. Then, use the diagrams to find the critical section and calculate the maximum bending stress in segment AC.

Problem 2

The torsion rod with variable cross-section shown in Fig. 2.1 is clamped at A and carries point torques at B (4 kN.m), C (8 kN.m) and D (unknown value T) with the senses indicated in the figure. It is known that the diameter of AB is 25 mm, the diameter of BC is 50 mm and the diameter of CD is 20 mm. Furthermore, 6LAB = 6LBC = 5LCD = 1200 mm. If the shear modulus of the material is 80 GPa, find:

a) The value of T that would make the angle of twist at point C with respect to A equal to zero.

b) The maximum value of T that can applied with the sense shown such that failure does not occur for an allowable shear stress of 1.1 GPa. Can the condition from part a be achieved without failure?

Fig. 2.1: Torsion rod with variable cross-section.

Problem 3

The cantilever beam shown in Fig. 3.1 has a rectangular cross-section with h = 120 mm and b = 80 mm. The 12 kN and 10 kN forces act parallel to the x and z axis, respectively, and pass through the centroid of the beam cross-section at the locations they act. The 10 kN force acts at the free end of the cantilever, whereas the 12 kN force acts 250 mm away from the free end. The cross-section ABCD is 750 mm away from the free end.

a) Determine the magnitude and location of the maximum tensile and compressive bending stress at the cross-section ABCD and indicate the neutral axis on it.

b) If an additional force, F, is applied on the beam parallel to the z axis at a point 500 mm away from the free end, what should be its magnitude and direction to make the bending stress at point B zero?

Fig. 3.1: Cantilever beam with loads along two axes.

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College Physics by Openstax Chapter 6 Problem 28

Riding a Bicycle in an Ideally Banked Curve


Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that \theta (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, \theta = \tan ^{-1} \left( v^2/rg \right)

(b) Calculate \theta for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6.33 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system, while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force N and F_c are the vertical and horizontal components of the force F.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for F, we have

\begin{align*}
\sum F_y & = 0 \\ \\
F \cos \theta - mg & = 0 \\ \\
F \cos \theta & = mg \\ \\
F & = \frac{mg}{\cos \theta}  \quad \quad  & \color{Blue}  \small \text{Equation 1}
\end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

\begin{align*}
\sum F_x & = ma_c \\ \\
F \sin \theta  & = m a_c \\ \\
F \sin \theta  & = m \frac{v^2}{r}   \quad \quad  & \color{Blue}  \small \text{Equation 2}
\end{align*}

We substitute Equation 1 to Equation 2.

\begin{align*}
F \sin \theta  & = m \frac{v^2}{r} \\ \\
\frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\
mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\
\end{align*}

We can cancel m from both sides, and we can apply the trigonometric identity \displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

\begin{align*}
g \tan \theta & = \frac{v^2}{r} \\ \\
\tan \theta & = \frac{v^2}{rg} \\ \\
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following values:

  • linear velocity, v = 12.0\ \text{m/s}
  • radius of curvature, r=30.0\ \text{m}
  • acceleration due to gravity, g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of \theta we solve in Part A.

\begin{align*}
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\
\theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\
\theta & = 26.0723 ^\circ \\ \\
\theta & = 26.1 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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