f5ch13 - 356 CHAPTER 13 EQUILIBRIUM AND ELASTICITY CHAPTER...

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356 CHAPTER 13 EQUILIBRIUM AND ELASTICITY C HAPTER 13 Answer to Checkpoint Questions . c, e, f . ( a ) no; ( b ) at site of F , perpendicular to plane of gure; ( c ) N . ( a ) at C (to eliminate forces there from a torque equation); ( b ) plus; ( c ) minus; ( d ) equal . d . ( a ) equal; ( b ) B; ( c ) B Answer to Questions . ( a ) yes; ( b ) yes; ( c ) yes; ( d ) no . a and c (forces and torques balance) . b . m = kg, m = kg, m = kg . ( a ) yes; ( b ) no; ( c ) no (it could balance the torques but the forces would then be unbalanced) . ( a ) ; ( b ) ; ( c ) ; ( d ) tower built straight up, center of mass over center of support area . ( a ) a, then b and c tie, and then d . ( a ) same; ( b ) smaller; ( c ) smaller; ( d ) same . ( a ) N (the key is the pulley with the -N weight); ( b ) N . ( a ) N (the key is the pulley with the -N pi~nata) . ( a ) sin ; ( b ) same; ( c ) larger . A, then tie of B and C
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θ 55 m 7.0 m x= 7.0 m cm CHAPTER 13 EQUILIBRIUM AND ELASTICITY 357 . tie of A and B, then C E Use = mgr to calculate the torque exerted by the weight of a person with mass m sitting at a distance r from point f . Note that persons No. through exert torques pointing out of the page, and persons No. through exert torques pointing into the page. You can easily check that person No. exerts the greatest torque out of the page while person No. exerts the greatest torque into the page. E The force component required is F ? = ( N)( : cm = cm) = : N : E ( a ) The critical situation of the tower when it is on the verge of toppling is shown to the right. The top of the tower would be displaced : m from the vertical, requiring an additional : m on top of the present displacement. ( b ) The angle of inclination is = sin : m m = : : E ( a ) The forces are balanced when they sum to zero: F + F + F = . This means F = F F = ( i j ) N ( i + j ) N = ( i + j ) N. ( b ) If is the angle the vector makes with the x axis then tan = F y F x = N N = : : The angle is either : or . The second solution yields a negative x component and a positive y component and is therefore the correct solution.
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α θ T T F T T F l h m g T T M 3 g T 2 T 2 M 2 f s,max 358 CHAPTER 13 EQUILIBRIUM AND ELASTICITY E From the gure to the right we have T sin = F = T , which gives = sin ( = ) = . The angle between the two parts of the string is then = = . E The free-body diagram for the rope is shown to the right. The force T which the car exerts on the rope satis es T sin = F , which gives T = F sin = lb ( : ft = ft) = : lb : This is the same as the magnitude of the force exerted by the rope on the car. E The equilibrium of forces requires that T sin = mg; where tan h=l . Thus
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f5ch13 - 356 CHAPTER 13 EQUILIBRIUM AND ELASTICITY CHAPTER...

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