PSE9e ISM chapter12 final tủ tài liệu training

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PSE9e ISM chapter12 final tủ tài liệu training

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12 Static Equilibrium and Elasticity CHAPTER OUTLINE 12.1 Analysis Model: Rigid Object in Equilibrium 12.2 More on the Center of Gravity 12.3 Examples of Rigid Objects in Static Equilibrium 12.4 Elastic Properties of Solids * An asterisk indicates an item new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ12.1 Answer (b) The skyscraper is about 300 m tall The gravitational field (acceleration) is weaker at the top by about 900 parts in ten million, by on the order of 10−4 times The top half of the uniform building is lighter than the bottom half by about (1/2)(10−4) times Relative to the center of mass at the geometric center, this effect moves the center of gravity down, by about (1/2)(10−4)(150 m) ~ 10 mm OQ12.2 Answer (c) Net torque = (50 N)(2 m) − (200 N)(5 m) − (300 N)x = 0; therefore, x = m OQ12.3 Answer (a) Our theory of rotational motion does not contradict our previous theory of translational motion The center of mass of the object moves as if the object were a particle, with all of the forces applied there This is true whether the object is starting to rotate or not OQ12.4 Answer (d) In order for an object to be in equilibrium, it must be in both translational equilibrium and rotational equilibrium Thus, it  must meet two conditions of equilibrium, namely F net = and  τ net = 632 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 633 OQ12.5 Answer (b) The lower the center of gravity, the more stable the can In cases (a) and (c) the center of gravity is above the base by one-half the height of the can In case (b), the center of gravity is above the base by only a bit more than one-quarter of the height of the can OQ12.6 Answer (d) Using the left end of the plank as a pivot and requiring that ∑ τ = gives −mg ( 2.00 m ) + F2 ( 3.00 m ) = or ( ) 2mg ( 20.0 kg ) 9.80 m s F2 = = = 131 N 3 OQ12.7 Answer: τD > τC > τE > τB > τA The force exerts a counterclockwise torque about pivot D The line of action of the force passes through C, so the torque about this axis is zero In order of increasing negative (clockwise) values come the torques about F, E and B essentially together, and A OQ12.8 Answer (e) In the problems we study, the forces applied to the object lie in a plane, and the axis we choose is a line perpendicular to this plane, so it appears as a point on the force diagram It can be chosen anywhere The algebra of solving for unknown forces is generally easier if we choose the axis where some unknown forces are acting OQ12.9 (i) Answer (b) The extension is directly proportional to the original dimension, according to F/A = Y∆L/Li (ii) Answer (e) Doubling the diameter quadruples the area to make the extension four times smaller OQ12.10 Answer (b) Visualize the ax as like a balanced playground seesaw with one large-mass person on one side, close to the fulcrum, and a small-mass person far from the fulcrum on the other side Different masses are on the two sides of the center of mass The mean position of mass is not the median position © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 634 Static Equilibrium and Elasticity ANSWERS TO CONCEPTUAL QUESTIONS CQ12.1 The free-body diagram demonstrates that it is necessary to have friction on the ground to counterbalance the normal force of the wall and to keep the base of the ladder from sliding If there is friction on the floor and on the wall, it is not possible to determine whether the ladder will slip, from the equilibrium conditions alone CQ12.2 A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide at the edge of a swimming pool each have a center of mass that is not within the bulk of the object CQ12.3 ANS FIG CQ12.1 (a) Consider pushing up with one hand on one side of a steering wheel and pulling down equally hard with the other hand on the other side A pair of equal-magnitude oppositely-directed forces applied at different points is called a couple (b) An object in free fall has a nonzero net force acting on it, but a net torque of zero about its center of mass CQ12.4 When one is away from a wall and leans over, one’s back moves backward so the body’s center of gravity stays over the feet When standing against a wall and leaning over, the wall prevents the backside from moving backward, so the center of gravity shifts forward Once your CG is no longer over your feet, gravity contributes to a nonzero net torque on your body and you begin to rotate CQ12.5 If an object is suspended from some point and allowed to freely rotate, the object’s weight will cause a torque about that point unless the line of action of its weight passes through the point of support Suspend the plywood from the nail, and hang the plumb bob from the nail Trace on the plywood along the string of the plumb bob The plywood’s center of gravity is somewhere along that line Now suspend the plywood with the nail through a different point on the plywood, not along the first line you drew Again hang the plumb bob from the nail and trace along the string The center of gravity is located halfway through the thickness of the plywood under the intersection of the two lines you drew CQ12.6 She can be correct Consider the case of a bridge supported at both ends: the sum of the forces on the ends equals the total weight of the bridge If the dog stands on a relatively thick scale, the dog’s legs on © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 635 the ground might support more of its weight than its legs on the scale She can check for and if necessary correct for this error by having the dog stand like a bridge with two legs on the scale and two on a book of equal thickness—a physics textbook is a good choice CQ12.7 Yes, it can Consider an object on a spring oscillating back and forth In the center of the motion both the sum of the torques and the sum of the forces acting on the object are (separately) zero Again, a meteoroid flying freely through interstellar space feels essentially no forces and keeps moving with constant velocity CQ12.8 Shear deformation Its deformations are parallel to its surface SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 12.1 P12.1 Analysis Model: Rigid Object in Equilibrium Use distances, angles, and forces as shown in ANS FIG P12.1 The conditions of equilibrium are: ∑ Fy = ⇒ Fy + Ry − Fg = ∑ Fx = ⇒ Fx − Rx = ⎛ ⎞ ∑ τ = ⇒ Fy cosθ − Fg ⎜⎝ ⎟⎠ cosθ − Fx sin θ = ANS FIG P12.1 P12.2 Take torques about P, as shown in ANS FIG P12.2   ⎡ ⎤ ⎡ ⎤ ∑ τ p = −nO ⎢ + d ⎥ + m1 g ⎢ + d ⎥ + mb gd − m2 gx = ⎣2 ⎦ ⎣2 ⎦ We want to find x for which nO = 0: x= ( m1 g + mb g ) d + m1 g 2 ( m1 + mb ) d + m1 2 m2 g = m2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 636 Static Equilibrium and Elasticity For the values given: x= ( m1 + mb ) d + m1 2 x= ( 5.00 kg + 3.00 kg )( 0.300 m ) + ( 5.00 kg ) 1.002 m m2 15.0 m x = 0.327 m ANS FIG P12.2 The situation is impossible because x is larger than the remaining portion of the beam, which is 0.200 m long Section 12.2 P12.3 More on the Center of Gravity The coordinates of the center of gravity of piece are x1 = 2.00 cm and y1 = 9.00 cm The coordinates for piece are x2 = 8.00 cm and y = 2.00 cm The area of each piece is A1 = 72.0 cm and A2 = 32.0 cm ANS FIG P12.3 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 637 And the mass of each piece is proportional to the area Thus, xCG = 2 ∑ mi xi = ( 72.0 cm )( 2.00 cm ) + ( 32.0 cm )( 8.00 cm ) 72.0 cm + 32.0 cm ∑ mi = 3.85 cm and 2 mi y i ( 72.0 cm )( 9.00 cm ) + ( 32.0 cm )( 2.00 cm ) ∑ y CG = = 104 cm ∑ mi = 6.85 cm P12.4 The definition of the center of gravity as the average position of mass in the set of objects will result in equations about x and y coordinates that we can rearrange and solve to find where the last mass must be     mi ri ∑ , r CG( ∑ mi ) = ∑ mi ri From rCG = ∑ mi We require the center of mass to be at the origin; this simplifies the equation, leaving ∑ mi xi = and ∑ mi yi = To find the x coordinate, we substitute the known values: ( 5.00 kg )( m ) + ( 3.00 kg )( m ) + ( 4.00 kg )( 3.00 m ) + ( 8.00 kg ) x = Solving for x gives x = –1.50 m Likewise, to find the y coordinate, we solve: ( 5.00 kg )( m ) + ( 3.00 kg )( 4.00 m ) + ( 4.00 kg )( m ) + ( 8.00 kg ) y = to find y = –1.50 m Therefore, a fourth mass of 8.00 kg should be located at  r4 = (–1.50ˆi − 1.50ˆj) m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 638 P12.5 Static Equilibrium and Elasticity Let σ represent the mass-per-face area (It would be equal to the material’s density multiplied by the constant thickness of the wood.) A ( x − 3.00 )2 has vertical strip at position x, with width dx and height , mass σ ( x − 3.00 ) dx dm = The total mass is M = ∫ dm = σ ( x − ) dx ⎛ σ ⎞ =⎜ ⎟ ∫x=0 ⎝ 9⎠ 3.00 ∫ (x − 6x + ) dx 3.00 ⎤ ⎛ σ ⎞ ⎡ x 6x =⎜ ⎟⎢ − + 9x ⎥ ⎝ 9⎠⎣ ⎦0 3.00 =σ The x coordinate of the center of gravity is xCG ∫ x dm = = M = 9σ 3.00 ∫ σ σ x ( x − ) dx = 9σ 3.00 ⎡ x 6x 9x ⎤ − + ⎢⎣ ⎥⎦ = 3.00 ∫ (x − 6x + 9x ) dx 6.75 m = 0.750 m 9.00 ANS FIG P12.5 P12.6 We can visualize this as a whole pizza with mass m1 and center of gravity located at x1, plus a hole that has negative mass, –m2, with center of gravity at x2: xCG = m1 x1 − m2 x2 m1 − m2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 639 Call σ the mass of each unit of pizza area R −R ⎞ σπ R − σπ ⎛ ⎞ ⎛ ⎝ 2⎠ ⎝ ⎠ xCG = R σπ R − σπ ⎛ ⎞ ⎝ 2⎠ xCG = P12.7 R/8 R = 3/4 In a uniform gravitational field, the center of mass and center of gravity of an object coincide Thus, the center of gravity of the triangle is located at x = 6.67 m, y = 2.33 m (see Example 9.12 on the center of mass of a triangle in Chapter 9) The coordinates of the center of gravity of the three-object system are then: xCG = = = y CG = = = Section 12.3 P12.8 ∑ mi xi ∑ mi (6.00 kg )( 5.50 m ) + ( 3.00 kg )(6.67 m ) + ( 5.00 kg )( −3.50 m ) ( 6.00 + 3.00 + 5.00 ) kg 35.5 kg ⋅ m = 2.54 m and 14.0 kg ∑ mi yi ∑ mi (6.00 kg )(7.00 m ) + ( 3.00 kg )( 2.33 m ) + ( 5.00 kg )( +3.50 m ) 14.0 kg 66.5 kg ⋅ m = 4.75 m 14.0 kg Examples of Rigid Objects in Static Equilibrium The car’s weight is Fg = mg = ( 500 kg ) ( 9.80 m/s ) = 4700 N  Call F the force ofthe ground on each of the front wheels and R the normal force on each of the rear wheels If we take torques around the front axle, with counterclockwise in the picture ANS FIG P12.8 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 640 Static Equilibrium and Elasticity chosen as positive, the equations are as follows: ∑ Fx = 0: = ∑ Fy = 0: 2R – 14 700N + 2F = ∑ τ = 0: +2R(3.00m) – (14 700 N)(1.20m) + 2F(0) = The torque equation gives: R= 17 640 N ⋅ m = 940 N = 2.94 kN 6.00 m Then, from the second force equation, 2(2.94 kN) – 14.7 kN + 2F = P12.9 and F = 4.41 kN The second condition for equilibrium at the pulley is [1] ∑ τ = = mg ( 3r ) − Tr and from equilibrium at the truck, we obtain 2T − Mg sin 45.0° = Mg sin 45.0° T= (1 500 kg ) g sin 45.0° = = 530g N ANS FIG P12.9 solving for the mass of the counterweight from [1] and substituting gives m= P12.10 (a) T 530g = = 177 kg 3g 3g For rotational equilibrium of the lowest rod about its point of support, ∑ τ = + ( 12.0 g ) g ( 3.00 cm ) − m1 g ( 4.00 cm ) = which gives m1 = 9.00 g (b) For the middle rod, + m2 g ( 2.00 cm ) − ( 12.0 g + 9.0 g ) g ( 5.00 cm ) = © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 641 which gives m2 = 52.5 g (c) For the top rod, ( 52.5 g + 12.0 g + 9.0 g ) g ( 4.00 cm ) − m g (6.00 cm ) = which gives m3 = 49.0 g P12.11 Since the beam is in equilibrium, we choose the center as our pivot point and require that ∑ τ center = −FSam ( 2.80 m ) + FJoe ( 1.80 m ) = or FJoe = 1.56FSam [1] ∑ Fy = ⇒ FSam + FJoe = 450 N [2] Also, Substitute equation [1] into [2] to get the following: FSam + 1.56FSam = 450 N or FSam = 450 N = 176 N 2.56 Then, equation [1] yields FJoe = 1.56 ( 176 N ) = 274 N Sam exerts an upward force of 176 N Joe exerts an upward force of 274 N P12.12 (a) To find U, measure distances and forces from point A Then, balancing torques, ( 0.750 m )U = ( 29.4 N )( 2.25 ) (b) U = 88.2 N To find D, measure distances and forces from point B Then, balancing torques, ( 0.750 m ) D = ( 1.50 m )( 29.4 N ) D = 58.8 N Also, notice that U = D + Fg , so ∑ Fy = P12.13 (a) The wall is frictionless, but it does exert a horizontal normal force, nw ∑ Fx = f − nw = ∑ Fy = ng − 800 N − 500 N = © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 P12.54 (a) 669 The height of pin B is (10.0 m ) sin 30.0° = 5.00 m The length of bar BC is then BC = 5.00 m = 7.07 m sin 45.0° ANS FIG P12.54(a) Consider the entire truss: ∑ Fy = nA − 000 N + nC = ∑ τ A = − ( 000 N )10.0cos 30.0° + nC [10.0cos 30.0° + 7.07 cos 45.0° ] = Which gives nC = 634 N Then, nA = 000 N − nC = 366 N (b) Joint A: ∑ Fy = 0: −CAB sin 30.0° + 366 N = so CAB = 732 N ∑ Fx = 0: −CAB cos 30.0° + TAC = TAC = ( 732 N ) cos 30.0° = 634 N ANS FIG P12.54(b) Joint B: ∑ Fx = 0: CBC = P12.55 ( 732 N ) cos 30.0° − CBC cos 45.0° = (732 N ) cos 30.0° = cos 45.0° 897 N Considering the torques about the point at the bottom of the bracket yields: W ( 0.050 m ) − Fhor ( 0.060 m ) = so Fhor = 0.833W (a) With W = 80.0 N, Fhor = 0.833 ( 80 N ) = 66.7 N (b) Differentiate with respect to time: dFhor /dt = 0.833 dW/dt Given that dW/dt = 0.150 N/s: The force exerted by the screw is increasing at the rate dFhor /dt = 0.833(0.150 N/s) = 0.125 N/s © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 670 P12.56 Static Equilibrium and Elasticity Refer to the solution to P12.57 for a general discussion of the solution From the geometry of the ladder, observe that cos θ = → θ = 75.5° In the following, we use the variables m = 70.0 kg, length AC = BC =  = 4.00 m, and d = 3.00 m ANS FIG P12.56 Consider the net torque about point A (on the bottom left side of the ladder) from external forces on the whole ladder The torques about A come from the weight of the painter and the normal force nB  ∑ τ A = −mgd cos 75.5° + nB = → nB = 2 mgd ⎛ 1⎞ mgd cos 75.5° = mgd ⎜ ⎟ → nB = ⎝ 4⎠   2 Consider the net torque about point B (on the bottom right side of the ladder) from external forces on the whole ladder The torques about B come from the weight of the painter and the normal force nA   2  ⎛ ⎞ → nA = mg ⎜ − d cos 75.5°⎟ ⎝ ⎠ 2 d⎞ ⎛ ⎛ ⎞ nA = mg ⎜ − d cos 75.5°⎟ = mg ⎜ − ⎟ ⎝2 ⎠ ⎝  2 ⎠ ⎛ ⎞ ∑ τ B = −nA + mg ⎜⎝ − d cos 75.5°⎟⎠ = Consider the torque from external forces about point C at the top of the right half of the ladder:   mgd → T = nB = sin 75.5° 2 sin 75.5° ∑ τ C = −T sin 75.5° + nB = →T = mgd 4 sin 75.5° Note that the tension T on the right half of the ladder must pull to the left, otherwise it could not contribute a clockwise torque about C to balance the counterclockwise torque from nB © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 671 Now we find the components of the reaction force that the left half of the ladder exerts on the right half Consider the forces acting on the right half of the ladder: ∑ Fx = Rx − T = → Rx = T, to the right ∑ Fy = Ry + nB = → Ry = −nB → Ry = nB , downward Collecting our results, we find (a) 70.0 kg ) ( 9.80 m/s )( 3.00 m ) ( mgd T= = → 4 sin 75.5° ( 4.00 m ) sin 75.5° (b) ⎛ d⎞ nA = mg ⎜ − ⎟ 2 ⎠ ⎝ T = 133 N ⎛ ( 3.00 m ) ( ) ⎞ nA = ( 70.0 kg ) 9.80 m/s ⎜ − ⎟ → nA = 429 N 4.00 m ⎝ ⎠ ( ) and ) ( mgd ( 70.0 kg ) 9.80 m/s ( 3.00 m ) nB = = → n = 257 N B 2 ( 4.00 m ) (c) The force exerted by the left half of the ladder on the right half is to the right and downward: Rx = T → Rx = 133 N, to the right and Ry = −nb → Ry = −257 N → Ry = 257 N, downward P12.57 From the geometry of Figure P12.56 and ANS FIG P12.56, we observe that cos θ = 4 =  and 15 ⎛ 1⎞ sin θ = − cos θ = − ⎜ ⎟ = − = ⎝ 4⎠ 16 16 sin θ = (a) 15 Below in part (b) we show that normal force nB = mgd/2 We use this result here to find the tension T in the horizontal bar © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 672 Static Equilibrium and Elasticity Consider the torque about point C at the top of the right half of the ladder:   mgd ⎛  ⎞ mgd mgd ⎛ ⎞ T = nB ⎜ ⎟ = = = ⎜ ⎟ ⎝ ⎠  sin θ 2 ⎝ ⎠  sin θ 4 sin θ 4 15 ∑ τ C = −T sin θ + nB = ( T= ) mgd  15 Note that the tension T on the right half of the ladder must pull to the left, otherwise it could not contribute a clockwise torque about C to balance the counterclockwise torque from nB (b) We now proceed to find the normal forces nA and nB First, consider the net torque from all forces acting on the ladder about point B at the bottom right side of the whole ladder Note that tension T on the left half of the ladder and tension T on the right half of the ladder have opposite torques because they have the same moment arms about point B, so their torques cancel (they are forces internal to the system, so they cannot contribute to net torque) In like manner, torques from Rx and Ry on both halves of the ladder cancel in pairs (again, they are internal forces) The only contributing torques come from the weight of the painter and the normal force nA (these are forces external to the ladder)   2  ⎛  d⎞ nA = mg ⎜ − ⎟ ⎝ 4⎠  ⎛ 2 − d ⎞ nA = mg ⎜ ⎝ ⎟⎠ 2 ⎛ 2 − d ⎞ nA = mg ⎜ ⎝ ⎟⎠  ⎛ ⎞ ∑ τ B = −nA + mg ⎜⎝ − d cos θ ⎟⎠ = nA = mg ( 2 − d ) 2 Now, consider the net torque from all forces acting on the ladder about point A on the bottom left side of the whole ladder Similarly to the case of the torques about point B, the only contributing torques about A come from the weight of the painter © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 673 and the normal force nB (again, these are external forces)  ∑ τ A = −mgd cosθ + nB = → nB = (c) 2 mgd ⎛ 1⎞ mgd cosθ = mgd ⎜ ⎟ → nB = ⎝ ⎠   2 Now we find the components of the reaction force that the left half of the ladder exerts on the right half Consider the forces acting on the right half of the ladder: ∑ Fx = Rx − T = → Rx = T Rx = mgd , to the right 15 ∑ Fy = Ry + nB = → Ry = −nB = − Ry = P12.58 mgd , downward 2 (a) ( 10.0 − 1.00 ) m s ⎛ Δv ⎞ F = m ⎜ ⎟ = ( 1.00 kg ) = 500 N ⎝ Δt ⎠ 0.002 s (b) stress = (c) P12.59 mgd 2 (a) F 500 N = = 4.50 × 106 N m A ( 0.010 m ) ( 0.100 m ) Yes This is more than sufficient to break the board Take both balls together Their weight is 2mg = 3.33 N and their CG is at their contact point ∑ Fx = 0:+ P3 − P1 = → P3 = P1 ∑ Fy = 0: + P2 − 2mg = → P2 = 2mg = 3.33 N For torque about the contact point (CP) between the balls: ∑ τ CP = 0: P1 ( R cos 45.0° ) − P2 ( R cos 45.0° ) + P3 ( R cos 45.0° ) − mg ( R cos 45.0° ) + mg ( R cos 45.0° ) = → P1 − P2 + P3 = → P1 + P3 = P2 Substituting P3 = P1, we find 2P1 = P2 = 2mg → P1 = mg © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 674 Static Equilibrium and Elasticity Therefore, P1 = P3 = 1.67 N ANS FIG P12.59(a) (b) Take the upper ball The lines of action of its weight, of P1 , and of the normal force n exerted by the lower ball all go through its center, so for rotational equilibrium there can be no frictional force ∑ Fx = 0: ncos 45.0° − P1 = n= 1.67 N = 2.36 N cos 45.0° ∑ Fy = 0: nsin 45.0° − 1.67 N = gives the same result ANS FIG P12.59(b) P12.60 We will let F represent some stretching force and use algebra to combine the Hooke’s-law account of the stretching with the Young’smodulus account Then integration will reveal the work done as the wire extends  |F|= kΔL (a) According to Hooke’s law, Young’s modulus is defined as Y = F/A ΔL/L © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 675 By substitution, Y=k (b) L A k= or YA L The spring exerts force –kx The outside agent stretching it exerts force +kx We can determine the work done by integrating the force kx over the distance we stretch the wire W = −∫ ΔL ΔL F dx = −∫ (−kx)dx = ( ) x = ΔL YA Δ L YA ⎤ x dx = ⎡⎢ x ⎥ L ∫0 ⎣ L ⎦x=0 Therefore, W= P12.61 YA ( ΔL ) / L Let θ represent the angle of the wire with the vertical The radius of the circle of motion is r = L sin θ , where L = 0.850 m For the mass: v2 = mrω r T sin θ = m[ L sin θ ]ω ∑ Fr = mar = m Further, ANS FIG P12.61 T ΔL ΔL or T = AY ⋅ =Y⋅ A L L Thus, AY ⋅ ( ΔL L ) = mLω , giving ω= or P12.62 AY ⋅ ( ΔL L ) mL π ( 3.90 × 10−4 m ) ( 7.00 × 1010 N m ) ( 1.00 × 10−3 ) = (1.20 kg )( 0.850 m ) ω = 5.73 rad s (a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet ∑ τ = ⇒ −F ( 1.00 m ) + ( 400 N )( 0.300 m ) = yielding F = ( 400 N )( 0.300 m ) = 1.00 m 120 N ∑ Fx = ⇒ − f + 120 N = 0, or f = 120 N © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 676 Static Equilibrium and Elasticity ∑ Fy = ⇒ −400 N + n = so n = 400 N Thus, µs = (c) f 120 N = = 0.300 n 400 N Apply F’ at the upper rear corner and directed so θ + φ = 90.0° to obtain the largest possible lever arm ⎛ 1.00 m ⎞ θ = tan −1 ⎜ = 59.0° ⎝ 0.600 m ⎟⎠ Thus, φ = 90.0° − 59.0° = 31.0° Sum the torques about the lower front corner of the cabinet: so ANS FIG P12.62 − F′ (1.00 m )2 + ( 0.600 m )2 + ( 400 N ) ( 0.300 m ) = F′ = 120 N ⋅ m = 103 N 1.17 m Therefore, the minimum force required to tip the cabinet is 103 N applied at 31.0° above the horizontal at the upper left corner *P12.63 (a) Consider the torques about an axis perpendicular to the page through the left end of the rod, as shown in ANS FIG P12.63 ∑ τ = 0: T ( 6.00 m ) cos 30.0° − ( 100 N )( 3.00 m ) − ( 500 N )( 4.00 m ) = then, T= ( 100 N )( 3.00 m ) + ( 500 N )( 4.00 m ) ( 6.00 m ) cos 30.0° = 443 N (b) From the first condition for equilibrium, ∑ Fx = 0: Rx = T sin 30.0° = ( 443 N ) sin 30.0° = 221 N toward the right © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 677 Similarly, ∑ Fy = 0: Ry + T cos 30.0° − 100 N − 500 N = which gives Ry = 600 N − T cos 30.0° = 600 N − ( 443 N ) cos 30.0° = 217 N upward ANS FIG P12.63 P12.64 Let the original length (when the cable is laid horizontally on a frictionless surface) of an infinitesimal piece of the cable be dy Let the extension of this piece be dL when the cable is vertically Then, for the entire cable, ΔL =  ∫ dL  =  ∫ F dy AY where F is the weight of the cable below a point at position y Evaluating F, with µ as the mass per unit length, ΔL =  ∫ =  ( µ y ) g dy  =  µ g L y dy   AY ∫ AY ANS FIG P12.64 µ g ⎛ L2 ⎞ ⎛ µ gL2 ⎞  =  AY ⎜⎝ ⎟⎠ ⎜⎝ AY ⎟⎠ 2 ⎡ ( 2.40 kg/m ) ( 9.80 m/s )( 500 m ) ⎤ ΔL = ⎢ ⎥ ⎢⎣ ( 2.00 × 1011 N/m ) ( 3.00 × 10−4 m ) ⎥⎦ = 0.049 m = 4.90 cm © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 678 Static Equilibrium and Elasticity Challenge Problems P12.65 With  as large as possible, n1 and n2 will both be large The equality sign in f2 ≤ µ s n2 will be true, but the less-than sign will apply in f1 < µ s n1 Take torques about the lower end of the pole ⎛1 ⎞ n2  cos θ + Fg ⎜ ⎟ cos θ − f2  sin θ = ⎝2 ⎠ Setting f2 = 0.576n2 , the torque equation becomes n2 ( − 0.576 tan θ ) + F =0 g Since n2 > , it is necessary that − 0.576 tan θ < ∴ tan θ > = 1.736 0.576 ∴ θ > 60.1° d 7.80 ft ∴ = < = 9.00 ft sin θ sin 60.1° ANS FIG P12.65 P12.66 Consider forces and torques on the beam ∑ Fx = 0: ∑ Fy = 0: ∑ τ = 0: (a) R cosθ − T cos 53° = R sin θ + T sin 53° − 800 N = (T sin 53°)( 8.00 m ) − ( 600 N ) d − ( 200 N )( 4.00 m ) = Suppressing units, we find T= 600d + 800 = 93.9d + 125, in N sin 53° © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 (b) 679 From substituting back, R cosθ = [ 93.9d + 125 ] cos 53.0° R sin θ = 800 N − [ 93.9d + 125 ] sin 53.0° Dividing, tan θ = R sin θ 800 N = − tan 53.0° + R cosθ ( 93.9d+125) cos 53.0° ⎛ 32 ⎞ tan θ = ⎜ − tan 53.0° ⎝ 3d + ⎟⎠ (c) To find R we can work out R cos θ + R sin θ = R From the expressions above for R cos θ and R sin θ , R = T cos 53° + T sin 53° − 600T sin 53° + ( 800 N ) R = T − 600T sin 53° + 640 000 R = ( 93.9d + 125 ) − 278 ( 93.9d + 125 ) + 640 000 R = ( 8.82 × 103 d − 9.65 × 10 d + 4.96 × 105 ) 12 (d) As d increases, T grows larger, θ decreases, and R decreases until about d = 5.4 m, then it increases Notes as d increases, the d term predominates P12.67 Imagine gradually increasing the force P This will make the force of static friction at the bottom increase, so that the normal force at the wall increases and the friction force at the wall can increase As P reaches its maximum value, the cylinder will turn clockwise microscopically to stress the welds at both contact points and make both forces of friction increase to their maximum values When it is on the verge of slipping, the cylinder is in equilibrium ANS FIG P12.67 ∑ Fx = 0: → f1 = n2 = µ s n1 and f2 = µ s n2 ∑ Fy = 0: → P + n1 + f2 = Fg ∑ τ = 0: → −PR + f1R + f2 R = → P = f1 + f2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 680 Static Equilibrium and Elasticity As P grows, so f1 and f2 Therefore, since µ s = f1 = P + n1 + and P= (a) n1 = Fg n1 n1 + = n1 4 5⎛ ⎞ n1 = Fg becomes P + ⎜ P⎟ = Fg or P = Fg 4⎝ ⎠ Therefore, P = P12.68 n1 n n and f2 = = 2 then So P + , Fg Just three forces act on the rod: forces perpendicular to the sides of the trough at A and B, and its weight The lines of action of the normal forces at A and B will intersect at a point above the rod so that those forces will have no torque about this point The rod’s weight will cause a torque about the point of intersection as in ANS FIG P12.68(a), and the rod will not be in equilibrium unless the center of the rod lies vertically below the intersection point, as in ANS FIG P12.68(b) All three forces must be concurrent Then the line of action of the weight is a diagonal of the rectangle formed by the two normal forces, and the rod’s center of gravity is vertically above the bottom of the trough ANS FIG P12.68(a) (b) In ANS FIG P12.68(b), AO cos 30.0° = BO cos60.0° and ⎛ cos 30.0° ⎞ L = AO + BO = AO + AO ⎜ ⎝ cos 60.0° ⎟⎠ AO = 2 L ⎛ cos 30° ⎞ 1+ ⎜ ⎝ cos60° ⎟⎠ 2 = L © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 So cos θ = 681 AO = and θ = 60.0° L ANS FIG P12.68(b) (c) Unstable If the rod is displaced slightly, it will slip until it lies along the left edge of the trough where its center of gravity will be lower © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 682 Static Equilibrium and Elasticity ANSWERS TO EVEN-NUMBERED PROBLEMS P12.2 The situation is impossible because x is larger than the remaining portion of the beam, which is 0.200 m long P12.4 x = −1.50 m; y = −1.50 m P12.6 R P12.10 2.94 kN; 4.41 kN P12.8 (a) m1 = 9.00 g; (b) m2 = 52.5 g; (c) m3 = 49.0 g P12.12 (a) U = 88.2 N; (b) D = 58.8 N P12.14 ⎡1 ⎤ ⎛ x⎞ ( m1 / + m2 d / L cot θ (a) ⎢ m1 g + ⎜ ⎟ m2 g ⎥ cot θ , (m1 + m2)g; (b) ⎝ L⎠ m1 + m2 ⎣2 ⎦ P12.16 (a) See ANS FIG P12.16; (b) ) mg cot θ ; (c) T = µsmg; (d) µ s = cot θ ; 2 (e) The ladder slips P12.18 (a) See ANS FIG P12.18; (b) 392 N; (c) 339 N to the right; (d) 0; (e) V = 0; (f) 392 N; (g) 339 N to the right; (h) The two solutions agree precisely They are equally accurate P12.20 (a) No time interval The horse’s feet lose contact with the drawbridge as soon as it begins to move; (b) 1.73 rad/s; (c) 2.22 rad/s; (d) 6.62 kN The force at the hinge is 4.72 ˆi + 6.62 ˆj kN ; (e) 59.1 kJ ) ( P12.22 (a) mg 2Rh − h2 ( R − h) cosθ − 2Rh − h2 sin θ ; ⎡ ⎤ 2Rh − h2 cosθ (b) and mg + ⎢ ⎥ ( R − h) cosθ − 2Rh − h2 sin θ ⎢⎣ (R − h)cosθ − 2Rh − h sin θ ⎥⎦ mg 2Rh − h2 cosθ P12.24 (a) See ANS FIG P12.24; (b) 218 N; (c) 72.4 N; (d) 2.41 m; (e) See P12.24(e) for full explanation P12.26 ~ cm P12.28 (a) 73.6 kN; (b) 2.50 mm P12.30 1.0 × 1011 N/m2 P12.32 1.65 ì 108 N/m2 P12.34 8.60 ì 104 m â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 683 P12.36 9.85 × 10–5 P12.38 (a) Rigid object in static equilibrium; (b) See ANS FIG P12.38; (c) The woman is at x = when n1 is greatest; (d) n1 = 0; (e) 1.42 × 103 N; (f) 5.64 m; (g) same as answer (f) P12.40 (a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm; (d) 2.40 mm; (e) 48.0 kN P12.42 θ = 21.2°; T = 1.68 kN; R = 2.34 kN P12.44 (a) See ANS FIG P12.44 for the force diagram and see P12.44(a) for a sample problem statement (b) The upper hinge exerts 410 N to the left and 442 N up The lower hinge exerts 410 N to the right P12.46 T = 1.46 kN; H = 1.33 kN; V = 2.58 kN P12.48 (a) 2.71 kN; (b) 2.65 kN; (c) You should lift “with your knees” rather than “with your back”; (d) In this situation, you can make the compressional force in your spine about ten times smaller by bending your knees and lifting with your back as straight as possible P12.50 The situation is impossible because the new technique would tip the cabinet over P12.52 209 N P12.54 (a) nC = 634 N, nA = 000 N – nC = 366 N; (b) CAB = 732 N, TAC = 634 N, and CBC = 897 N P12.56 (a) T = 133 N; (b) nA = 429 N, nB = 257 N; (c) Rx = 133 N, to the right, Ry = 257 N, downward P12.58 (a) 500 N; (b) 4.50 × 106 N/m2; (c) yes P12.60 (a) P12.62 (a and b) 120 N, 0.300; (c) 103 N applied at 31.0° above the horizontal at the upper left corner P12.64 4.90 cm P12.66 (a) 93.9d + 125, in N; (b) See P12.66(b) for full derivation; (c) See P12.66(c) for full derivation; (d) As d increases, T grows larger, θ decreases, and R decreases until about d = 5.4 m, then it increases Note as d increases, the d2 term predominates P12.68 (a) See P12.68(a) for the full explanation; (b) 60.0°; (c) unstable YA (ΔL)2 ; (b) YA L 2L © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

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