Chapter 11 Mastering Physics

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Chapter 11 Homewor k Assignm ent

Chapter 11 Homework Assignment Due: 11:00pm on Saturday, October 10, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy

Video Tutor: Balancing a Meter Stick First, launch the video below. video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.

Part A Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot to keep the meter stick in balance?

Hint 1. How to approach the problem. For the meter stick to be in equilibrium, the net torque on it must be zero. Torques about the fulcrum may be exerted by the mass hanging from the end of the stick and by the stick’s own weight. Use the condition that the net torque must be equal to zero to obtain a relationship involving 1) the distance between the left end of the stick and the fulcrum and 2) the distance between the center of mass of the stick and the fulcrum. These two distances must add up to a constant. You should get two equations that you can solve for the location of the fulcrum.

ANSWER: 90 cm (10 cm from the weight) 75 cm (25 cm from the weight) 10 cm 50 cm (in the middle) 25 cm

Correct Typesetting Typesetting math: 20%

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Chapter 11 Homewor k Assignm ent

A Bar Suspended by Two Wires A nonu nonuniform, niform, hor horizontal izontal bar of mass is suppor supported ted by two massless wire wires s again against st gra gravity vity.. The left wire makes an angle with the horizontal, and the right wire makes an angle . The bar has length .

Part A Find the position position of the center of mass of the bar, bar, Express the center of mass in terms of

,

, measured from the bar's left end.

, and

.

Hint 1. Nature of the problem This is a statics problem. There is no net force or torque acting on the bar.

Hint 2. Torques about left end of bar The net torque is zero about any point you select. Here we ask you to find the net torque of the system about the left end end of the bar. Label the tensi tension on in the left wire wire , and label the other wire's tensi tension on . The weight weig ht of the bar is . Note that the vector sum of , , and is zero. Using the sign convention shown in the picture, express the sum of the torques about the left end of the bar. Answer in terms of Answer in your answer.

,

,

,

,

,

, and/or

. Not Note e that not all of thes these e quantities will appear

ANSWER: =

Hint 3. Forces: x components components Assume that the tensions in the left and right wire wires s are and , respectively. Wha Whatt is the sum of the x components compone nts of the forces ? Because this is a static statics s pro problem, blem, these forces will sum to zero. Use the sign convention indi indicated cated in the figure, and express your answer in terms of , , and/or . Note that not all of these quantities will appear in your answer. https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnm entID= 3673484

,

,

,

,

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Chapter 11 Homework Assignment

ANSWER: =

Hint 4. Forces: y components Assuming that the tensions in the left and right wires are and , respectively, what is the sum of the y components of the forces ? Because this is a statics problem, these forces will sum to zero. Use the sign convention indicated in the figure, and express your answer in terms of , , and/or . Note that not all of these quantities will appear in your answer.

,

,

,

,

ANSWER: =

Hint 5. Eliminate weight from your equations You should have found three equations by now. It is possible to eliminate two variables and solve for in terms of the others. As an intermediate step, solve your torque equation for in terms of , \texttip{T_{\rm 2}}{T_2}, \texttip{L}{L}, etc. and then solve your y -component force equation for \texttip{W}{W} and substitute back into your expression for \texttip{x}{x}. In other words, find an expression for \texttip{x}{x}. Answer in terms of \texttip{T_{\rm 1}}{T_1}, \texttip{T_{\rm 2}}{T_2}, \texttip{\phi _{\rm 1}}{phi_1}, \texttip{\phi _{\rm 2}}{phi_2}, and \texttip{L}{L}. ANSWER: \texttip{x}{x} \large{\frac{T_{2} {\sin}\left({\phi}_{2}\right) L}{T_{1} {\sin}\left({\phi}_{1}\right)+T_{2} = {\sin}\left({\phi}_{2}\right)}}

Hint 6. A useful trig identity The dimensions for the expression you just found for \texttip{x}{x} are correct, since the units of the tensions cancel out, leaving the units of length in the numerator. If you now solve the x -component force equation for \texttip{T_{\rm 1}}{T_1} in terms of \texttip{T_{\rm 2}}{T_2}, and substitute into your equation for \texttip{x}{x}, you should find the following trig identity useful: \sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b). Alternatively, you could express your answer in terms of \tan(\phi_1) and \tan(\phi_2).

ANSWER: \texttip{x}{x} = \large{L {\frac{{\tan}\left({\phi}_{2}\right)}{{\tan}\left({\phi}_{2}\right)+{\tan}\left({\phi}_{1}\right)}}}

Correct

Precarious Lunch https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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A uniform steel beam of length \texttip{L}{L} and mass \texttip{m_{\rm 1}}{m_1} is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle \texttip{\theta }{theta}, as shown. Through the hinge, the wall exerts an unknown force, \texttip{F}{F}, on the beam. A workman of mass \texttip{m_{\rm 2}}{m_2} sits eating lunch a distance \texttip{d}{d} from the building.

Part A Find \texttip{T}{T}, the tension in the cable. Remember to account for all the forces in the problem. Express your answer in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{L}{L}, \texttip{d} {d}, \texttip{\theta }{theta}, and \texttip{g}{g}, the magnitude of the acceleration due to gravity.

Hint 1. Pick the best origin This is a statics problem so the sum of torques about any axis a will be zero. In order to solve for \texttip{T} {T}, you want to pick the axis such that \texttip{T}{T} will give a torque, but as few as possible other unknown forces will enter the equations. So where should you place the origin for the purpose of calculating torques? ANSWER: At the center of the bar At the hinge At the connection of the cable and the bar Where the man is eating lunch

Hint 2. Calculate the sum torques Now find the sum of the torques about center of the hinge. Remember that a positive torque will tend to rotate objects counterclockwise around the origin. Answer in terms of \texttip{T}{T}, \texttip{L}{L}, \texttip{d}{d}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{\theta }{theta}, and \texttip{g}{g}.

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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ANSWER: \Sigma\;\tau_a = 0 = \large{\left(\frac{m_{1} L}{2}+m_{2} d\right) g-T L {\sin}\left({\theta}\right)}

ANSWER: \texttip{T}{T} = \large{\frac{g\left(m_{1} {\frac{L}{2}}+m_{2} d\right)}{L {\sin}\left({\theta}\right)}}

Correct

Part B Find \texttip{F_{\mit x}}{F_x}, the \texttip{x}{x}-component of the force exerted by the wall on the beam ( \texttip{F} {F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T} and other given quantities.

Hint 1. Find the sign of the force The beam is not accelerating in the \texttip{x}{x}-direction, so the sum of the forces in the \texttip{x}{x}direction is zero. Using the given coordinate system, is \texttip{F_{\mit x}}{F_x} going to have to be positive or negative?

ANSWER: \texttip{F_{\mit x}}{F_x} = -T {\cos}\left({\theta}\right)

Correct

Part C Find \texttip{F_{\mit y}}{F_y}, the y-component of force that the wall exerts on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T}, \texttip{\theta }{theta}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. ANSWER: \texttip{F_{\mit y}}{F_y} = g\left(m_{1}+m_{2}\right)-T {\sin}\left({\theta}\right)


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Correct If you use your result from part (A) in your expression for part (C), you'll notice that the result simplifies somewhat. The simplified result should show that the further the luncher moves out on the beam, the lower the magnitude of the upward force the wall exerts on the beam. Does this agree with your intuition?

Three-Legged Table The top view of a table, with weight \texttip{W_{\rm t}}{W_t}, is shown in the figure. The table has lost the leg at (\texttip{L_{\mit x}}{L_x}, \texttip{L_{\mit y}}{L_y}), in the upper right corner of the diagram, and is in danger of tipping over. Company is about to arrive, so the host tries to stabilize the table by placing a heavy vase (represented by the green circle) of weight \texttip{W_{\rm v }}{W_v} at ( \texttip{X}{X}, \texttip{Y}{Y}). Denote the magnitudes of the upward forces on the table due to the legs at (0, 0), ( \texttip{L_{\mit x}}{L_x}, 0), and (0, \texttip{L_{\mit y}}{L_y}) as \texttip{F_{\rm 0}}{F_0}, \texttip{F_{\mit x}}{F_x}, and \texttip{F_{\mit y}}{F_y}, respectively.

Part A Find \texttip{F_{\mit x}}{F_x}, the magnitude of the upward force on the table due to the leg at ( \texttip{L_{\mit x}} {L_x}, 0). Express the force in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{X}{X}, \texttip{Y}{Y}, \texttip{L_{\mit x}}{L_x}, and/or \texttip{L_{\mit y}}{L_y}. Note that not all of these quantities may appear in the answer.

Hint 1. Find the useful vector relations This is a statics problem, so about any point \texttip{p}{p} we have \sum \vec{\tau_{p}} = 0 and, at \texttip{p} {p}, \sum \vec{F} = 0. Each is a vector equation, yielding algebraic expressions for each of the three components, x , y , and z . Using the coordinate system shown in the figure, indicate for each component of these equations whether it gives a useful relationship among the various forces. Check all that apply.

Hint 1. Direction of applied forces All of the forces are in the z direction: Figure out what nonzero components of the torque these could possibly generate.

ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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x component of the forces y component of the forces z component of the forces x component of the torques y component of the torques z component of the torques

Hint 2. Find the y component of the torque What is \sum \tau_y, the y component of the torque equation? Express the y component of the torque in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{F_{\mit x}}{F_x}, \texttip{F_{\mit y}}{F_y}, \texttip{L_{\mit x}}{L_x}, \texttip{L_{\mit y}}{L_y}, \texttip{X}{X}, and/or \texttip{Y}{Y}. Note that not all terms may appear in the answer.

Hint 1. Choice of origin for torques Take torques around the origin of the coordinate system (i.e., this becomes point \texttip{p}{p} in the torque equations), as this will result in an equation with one and only one unknown force.

Hint 2. Sign convention If you do not get the sign from the cross-product expression \tau_y = z F_x - x F_z , remember that the y torque is positive for clockwise rotation about the y axis when looking in the positive y direction. This is a subtle point--normally you look down on the x -y axes from above, that is, toward negative z , and so positive torque (about the z axis) is counterclockwise.

ANSWER: \sum \tau_y=0 = \large{-W_{v} X+\left(F_{x}-\frac{W_{t}}{2}\right) L_{x}}

ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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\texttip{F_{\mit x}}{F_x} = \large{{\frac{W_{v} X}{L_{x}}}+{\frac{W_{t}}{2}}}

Correct

Part B Find \texttip{F_{\mit y}}{F_y}, the magnitude of the upward force on the table due to the leg at (0, \texttip{L_{\mit y}} {L_y}). Express the force in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{Y}{Y}, \texttip{X}{X}, \texttip{L_{\mit y}}{L_y}, and/or \texttip{L_{\mit x}}{L_x}. Note that not all of these quantities may appear in the final answer.

Hint 1. Find the x component of torque What is \sum \tau_x, the x component of the torque equation? Answer in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{F_{\mit x}}{F_x}, \texttip{F_{\mit y}}{F_y}, \texttip{L_{\mit x}}{L_x}, \texttip{L_{\mit y}}{L_y}, \texttip{X}{X}, and/or \texttip{Y}{Y}. Note that not all terms may appear in the answer. Remember to pay attention to the signs of the torque. ANSWER: \sum \tau_x=0 = \large{-W_{v} Y-\left(\frac{W_{t}}{2}-F_{y}\right) L_{y}}

ANSWER: \texttip{F_{\mit y}}{F_y} = \large{{\frac{W_{v} Y}{L_{y}}}+{\frac{W_{t}}{2}}}

Correct

Part C Find \texttip{F_{\rm 0}}{F_0}, the magnitude of the upward force on the table due to the leg at (0, 0). Express the force in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{L_{\mit x}}{L_x}, \texttip{L_{\mit y}}{L_y}, \texttip{X}{X}, and/or \texttip{Y}{Y}. Note that not all terms may appear in the answer.

Hint 1. Summing the total forces What is \sum F_z, the total vertical force on the table? Answer in terms of \texttip{W_{\rm v}}{W_v}, \texttip{W_{\rm t}}{W_t}, \texttip{F_{\mit x}}{F_x}, \texttip{F_{\mit y}}{F_y}, and/or \texttip{F_{\rm 0}}{F_0}. Remember to pay attention to signs (with a positive force being upward on the table).


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ANSWER: \sum F_{z}= 0 = F_{0}+F_{x}+F_{y}-W_{v}-W_{t}

ANSWER: \texttip{F_{\rm 0}}{F_0} = \large{W_{v} \left(1-{\frac{X}{L_{x}}}-{\frac{Y}{L_{y}}}\right)}

Correct

While the host is greeting the guests, the cat (of weight \texttip{W_{\rm c}}{W_c}) gets on the table and walks until her position is (x,y)=(L_x,L_y).

Part D Find the maximum weight \texttip{W_{\rm max}}{W_max} of the cat such that the table does not tip over and break the vase. Express the cat's weight in terms of \texttip{W_{\rm v}}{W_v}, \texttip{X}{X}, \texttip{Y}{Y}, \texttip{L_{\mit x}}{L_x}, and \texttip{L_{\mit y}}{L_y}.

Hint 1. Force when table begins to tip When the table is just beginning to tip over, what is the force \texttip{F_{\rm crit}}{F_crit} supplied by the leg at the origin?


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ANSWER: \texttip{F_{\rm crit}}{F_crit} = 0

Hint 2. Putting it all together Go back to your previous equations, put in the weight and/or torque from the weight of the cat (\texttip{W_{\rm c}}{W_c}) to obtain a new expression for \texttip{F_{\rm 0}}{F_0} (checking your signs carefully), equate this to the value you just obtained for \texttip{F_{\rm crit}}{F_crit}, and solve for \texttip{W_{\rm c}}{W_c} (now \texttip{W_{\rm max}}{W_max}).

ANSWER: \texttip{W_{\rm max}}{W_max} = \large{W_{v} \left(1-{\frac{X}{L_{x}}}-{\frac{Y}{L_{y}}}\right)}

Correct

Exercise 11.22 You are doing exercises on a Nautilus machine in a gym to strengthen your deltoid (shoulder) muscles. Your arms are raised vertically and can pivot around the shoulder joint, and you grasp the cable of the machine in your hand 64.0 {\rm {\rm cm}} from your shoulder joint. The deltoid muscle is attached to the humerus 15.0 {\rm {\rm cm}} from the shoulder oint and makes a 12.0 {\rm ^\circ} angle with that bone. (See the figure below .)


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Part A If you have set the tension in the cable of the machine to 36.0 {\rm {\rm N}} on each arm, what is the tension in each deltoid muscle if you simply hold your outstretched arms in place? ANSWER: T = 605

{\rm N}

Correct

Exercise 11.30 A vertical solid steel post of diameter \texttip{d}{d} = 22 {\rm {\rm cm}} and length \texttip{L}{L} = 2.40 {\rm {\rm m}} is required to support a load of mass \texttip{m}{m} = 7700 {\rm {\rm kg}} . You can ignore the weight of the post. Take free fall acceleration to be g=9.8 {\rm m/s^2}.

Part A What is the stress in the post? Express your answer using two significant figures. ANSWER: P = 2.0×106

{\rm Pa}

Correct

Part B https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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What is the strain in the post? Express your answer using two significant figures. ANSWER: \large{\frac {\Delta L} {L}} = 9.9×10−6

Correct

Part C What is the change in the post's length when the load is applied? Express your answer using two significant figures. ANSWER: \Delta L = 2.4×10−5

{\rm m}

Correct

Spinning Mass on a Spring An object of mass \texttip{M}{M} is attached to a spring with spring constant \texttip{k}{k} whose unstretched length is \texttip{L}{L}, and whose far end is fixed to a shaft that is rotating with angular speed \texttip{\omega }{omega}. Neglect gravity and assume that the mass rotates with angular speed \texttip{\omega }{omega} as shown. When solving this problem use an inertial coordinate system, as drawn here.

Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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Given the angular speed \texttip{\omega }{omega}, find the radius \texttip{R\left(\omega \right)}{R(omega)} at which the mass rotates without moving toward or away from the origin. Express the radius in terms of \texttip{k}{k}, \texttip{L}{L}, \texttip{M}{M}, and \texttip{\omega }{omega}.

Hint 1. Acceleration at a certain point Find \texttip{a\left(\omega \right)}{a(omega)}, the x component of the acceleration of mass \texttip{M}{M} at the instant pictured in the figure. Answer in terms of \texttip{R}{R} and \texttip{\omega }{omega}.

Hint 1. Expression for centripetal acceleration Recall that the magnitude of centripetal acceleration for constant circular motion is a_{\rm cent}=v^2/R. This expression can be converted into \texttip{\omega }{omega} and \texttip{R}{R} with the standard equation for velocity of distance/time.

ANSWER: \texttip{a\left(\omega \right)}{a(omega)} = R {\omega}^{2}

Hint 2. Graphical description of forces Which figure describes correctly the inertial force(s) acting on mass \texttip{M}{M}?

ANSWER: Figure 1 Figure 2 Figure 3 Figure 4

Hint 3. Spring force at a particular instant https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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What is \texttip{F_{\rm spring}\left(R\right)}{F_spring(R)}, the x component of the force exerted by the spring at the instant pictured in the figure? Answer in terms of \texttip{k}{k}, \texttip{R}{R}, and \texttip{L}{L}.

Hint 1. What determines the spring force? Recall that the amount of force a spring exerts is proportional to the distance it is stretched or compressed with respect to its equilibrium length ( \texttip{L}{L} in this case).

ANSWER: \texttip{F_{\rm spring}\left(R\right)}{F_spring(R)} = k \left(R-L\right)

Hint 4. Newton's 2nd law The mass is moving on a circular path, and so there is a constant change in the direction of its velocity vector and hence an inward radial acceleration. The vector sum of all forces acting on the object, \texttip{\vec{F}_{\rm total}}{F_total_vec}, must act to keep the mass on the circular trajectory at constant speed. Hence the forces are radial and F_{\rm total} = ma_{\rm radial}. Keep in mind that we are (as always) using an inertial reference system.

ANSWER: \texttip{R\left(\omega \right)}{R(omega)} = \large{\frac{kL}{k-M{\omega}^{2}}}

Correct

Part B Assume that, at a certain angular s peed \texttip{\omega _{\rm 2}}{omega_2}, the radius \texttip{R}{R} becomes twice \texttip{L}{L}. Find \texttip{\omega _{\rm 2}}{omega_2}. Answer in terms of \texttip{k}{k} and \texttip{M}{M}.

Hint 1. How to approach the problem In Part A, you have obtained the formula \large{R(\omega)=\frac{kL}{k-M\omega^2}}. Now use the additional piece of information: R(\omega_2)=2L to solve for \texttip{\omega _{\rm 2}}{omega_2}.

ANSWER: \texttip{\omega _{\rm 2}}{omega_2} = \large{\sqrt{\frac{k}{2M}}}


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Correct

Part C You probably have noticed that, as you increase \texttip{\omega }{omega}, there will be a value, \texttip{\omega _{\rm crit}}{omega_crit}, for which \texttip{R\left(\omega \right)}{R(omega)} goes to infinity. Find \texttip{\omega _{\rm crit}}{omega_crit}. Answer in terms of \texttip{k}{k} and \texttip{M}{M}.

Hint 1. What equation to use Note that the denominator of your expression for \texttip{R\left(\omega \right)}{R(omega)} is a function of \texttip{\omega }{omega}. The critical point(s) of \texttip{R\left(\omega \right)}{R(omega)} will coincide with the root(s) of the denominator. Before answering, analyze the physical validity of your answer.

ANSWER: \texttip{\omega _{\rm c rit}}{omega_crit} = \large{\sqrt{\frac{k}{M}}}

Correct

Part D What is happening to the spring as the angular velocity approaches \texttip{\omega _{\rm crit}}{omega_crit}? Choose the best option.

Hint 1. Graphical help Any spring behaves as shown in the two graphs:


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ANSWER: The spring streches linearly then breaks at \omega = \omega_{\rm crit}. The value of \texttip{\omega _{\rm crit}}{omega_crit} is so large that the spring will behave linearly for any practically attainable \texttip{\omega }{omega}. As \texttip{\omega }{omega} approaches \texttip{\omega _{\rm crit}}{omega_crit} the spring stops behaving linearly and begins to act more like an unstretchable rod until it eventually breaks.

Correct

± Young's Modulus Learning Goal: To understand the meaning of Young's modulus, to perform some real-life calculations related to stretching steel, a common construction material, and to introduce the concept of breaking stress. Hooke's law states that for springs and other "elastic" objects F=k\Delta x, where \texttip{F}{F} is the magnitude of the stretching force, \Delta x is the corresponding elongation of the spring from equilibrium, and \texttip{k}{k} is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, most materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be useful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length \texttip{L}{L} and cross-sectional area \texttip{A}{A} stressed by a force of magnitude \texttip{F}{F}. As a result, the bar stretches by \Delta L. Let us define two new terms: Tensile stress is the ratio of the stretching force to the cross-sectional area:

\large{{\rm stress}=\frac{F}{A}}. Tensile strain is the ratio of the elongation of the rod to the initial length of the bar: \large{{\rm strain}=\frac{\Delta L}{L}}. It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too large. That constant, which is an inherent property of a material, is called Young's modulus and is given by \large{Y=\frac{F/A}{\Delta L/L}.}

Part A What is the SI unit of Young's modulus?

Hint 1. Look at the dimensions If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension of https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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pressure. Use the SI unit of pressure.

ANSWER: Pa

Correct

Part B Consider a metal bar of initial length \texttip{L}{L} and cross-sectional area \texttip{A}{A}. The Young's modulus of the material of the bar is \texttip{Y}{Y}. Find the "spring constant" \texttip{k}{k} of such a bar for low values of tensile strain. Express your answer in terms of \texttip{Y}{Y}, \texttip{L}{L}, and \texttip{A}{A}.

Hint 1. Use the definition of Young's modulus Consider the equation defining \texttip{Y}{Y}. Then isolate \texttip{F}{F} and compare the result with Hooke's law: F = k\Delta x.

ANSWER: \texttip{k}{k} = \large{Y {\frac{A}{L}}}

Correct

Part C Ten identical steel wires have equal lengths \texttip{L}{L} and equal "spring constants" \texttip{k}{k}. The wires are connected end to end, so that the resultant wire has length 10L. What is the "spring constant" of the resulting wire?

Hint 1. The spring constant Use the expression for the spring constant determined in Part B. From the expression derived in the Part B, you can determine what happens to the spring constant when the length of the spring increases.

ANSWER: 0.1k \texttip{k}{k} 10k 100k


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Correct

Part D Ten identical steel wires have equal lengths \texttip{L}{L} and equal "spring constants" \texttip{k}{k}. The wires are slightly twisted together, so that the resultant wire has length \texttip{L}{L} and its cross-sectional area is ten times that of the individual wire. What is the "spring constant" of the resulting wire?

Hint 1. The spring constant Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can determine what happens to the spring constant when the area of the spring increases.

ANSWER: 0.1k \texttip{k}{k} 10k 100k

Correct

Part E Ten identical steel wires have equal lengths \texttip{L}{L} and equal "spring constants" \texttip{k}{k}. The Young's modulus of each wire is \texttip{Y}{Y}. The wires are connected end to end, so that the resultant wire has length 10L. What is the Young's modulus of the resulting wire? ANSWER: 0.1Y \texttip{Y}{Y} 10Y 100Y

Correct

Part F Ten identical steel wires have equal lengths \texttip{L}{L} and equal "spring constants" \texttip{k}{k}. The Young's modulus of each wire is \texttip{Y}{Y}. The wires are slightly twisted together, so that the resultant wire has length \texttip{L}{L} and is ten times as thick as the individual wire. What is the Young's modulus of the resulting wire? ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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0.1Y \texttip{Y}{Y} 10Y 100Y

Correct By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulus depends on the material, which remains unchanged. To change the Young's modulus, one would have to change the properties of the material itself, for instance by heating or cooling it.

Part G Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0 \times 10^{11} pascals. How far ( \texttip{\Delta L}{Delta L}) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters. ANSWER: \texttip{\Delta L}{Delta L} = 15

\rm {mm}

Correct Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, for greater values of tensile strain, the material no longer behaves elastically. If the strain and stress are large enough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the "stretching limit" of steel.

Part H Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0 \times 10^9 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second. Use two significant figures in your answer, expressed in kilometers.

Hint 1. Why does the cable break? The cable breaks because of the stress exerted on it by its own weight. At the moment that the breaking stress is reached, the stress at the top of the cable reaches its maximum, and the material begins to deteriorate. Introduce an arbitrary cross-sectional area of the cable (which will cancel out of the final answer). The mass of the cable below the top point can be found as the product of its volume and its density. Use this to find https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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the force at the top that will lead to the breaking stress.

Hint 2. Find the stress in the cable Assume that the cable has cross-sectional area \texttip{A}{A} and length \texttip{L}{L}. The density is \texttip{\rho }{rho}. The maximum stress in the cable is at the very top, where it has to support its own weight. What is this maximum stress? Express your answer in terms of \texttip{\rho }{rho}, \texttip{L}{L}, and \texttip{g}{g}, the magnitude of the acceleration due to gravity. Recall that the stress is the force per unit area, so the area will not appear in your expression. ANSWER: maximum stress =

{\rho} L g

ANSWER: 26

\rm {km}

Correct This is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is small enough to justify the assumption of virtually constant acceleration due to gravity. When making such assumptions, one should always check their validity after obtaining a result.

Problem 11.76 You are trying to raise a bicycle wheel of mass \texttip{m}{m} and radius \texttip{R}{R} up over a curb of height \texttip{h}{h}. To do this, you apply a horizontal force \texttip{\vec{F}}{F_vec}.

Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3673484

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What is the least magnitude of the force \texttip{\vec{F}}{F_vec} that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel? ANSWER: F = \large{\frac{mg \sqrt{2Rh-h^{2}}}{R-h}}

Correct

Part B What is the least magnitude of the force \texttip{\vec{F}}{F_vec} that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel? ANSWER: F = \large{mg \sqrt{\frac{h}{2R-h}}}

Correct

Part C In which case is less force required? ANSWER: case A case B

Correct

Problem 11.94 Compressive strength of our bones is important in everyday life. Young's modulus for bone is about 1.4 \times 10^{10}\; {\rm Pa}. Bone can take only about a 1.0% change in its length before fracturing.

Part A What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0\;{\rm cm}^{2}? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) Express your answer using two significant figures. ANSWER:


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F_{\rm max} = 4.2×104

{\rm N}

Correct

Part B Estimate the maximum height from which a 70{\rm -kg} man could jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be 0.006 {\rm s}, and assume that the stress is distributed equally between his legs. Express your answer using two significant figures. ANSWER: h_{\rm max} = 2.6

{\rm m}

Correct

Problem 11.83 A garage door is mounted on an overhead rail (the figure ). The wheels at A and B have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.49. The distance between the wheels is 2.00 {\rm m}, and each is 0.50 {\rm m} from the vertical sides of the door. The door is uniform and weighs 945 {\rm {\rm N}} . It is pushed to the left at constant speed by a horizontal force \vec F.

Part A If the distance h is 1.48 {\rm {\rm m}} , what is the vertical component of the force exerted on the wheel A by the track? ANSWER: F_V = 130

{\rm N}


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Correct

Part B If the distance h is 1.48 {\rm {\rm m}} , what is the vertical component of the force exerted on the wheel B by the track? ANSWER: F_V = 815

{\rm N}

Correct

Part C Find the maximum value h can have without causing one wheel to leave the track. ANSWER: h_{\rm max} = 2.04

{\rm m}

Correct Score Summary: Your score on this assignment is 101%. You received 11.06 out of a possible total of 11 points.


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Chapter 11 Mastering Physics

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