Static Equilibrium and Elasticity

Transcription

1 Static Equilibrium and Elasticit CHPTER OUTLINE. The Conditions for Equilibrium. More on the Center of Gravit.3 Eamples of Rigid Objects in Static Equilibrium.4 Elastic Properties of Solids NSWERS TO QUESTIONS *Q. The force eerts counterclockwise torque about pivot D. The line of action of the force passes through C, so the torque about this ais is zero. In order of increasing negative (clockwise) values come the torques about F, E and B essentiall together, and. The answer is then τ D > τ C > τ F > τ E τ B > τ Q. When ou bend over, our center of gravit shifts forward. Once our CG is no longer over our feet, gravit contributes to a nonzero net torque on our bod and ou begin to rotate. Q.3 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 (separatel) zero. gain, a meteoroid fling freel through interstellar space feels essentiall no forces and keeps moving with constant velocit. Q.4 (a) Consider pushing up with one hand on one side of a steering wheel and pulling down equall hard with the other hand on the other side. pair of equal-magnitude oppositeldirected forces applied at different points is called a couple. (b) n object in free fall has a non-zero net force acting on it, but a net torque of zero about its center of mass. *Q.5 nswer (a). Our theor of rotational motion does not contradict our previous theor 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. Q.6 V-shaped boomerang, a barstool, an empt 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. Q.7 Suspend the plwood from the nail, and hang the plumb bob from the nail. Trace on the plwood along the string of the plumb bob. Now suspend the plwood with the nail through a different point on the plwood, not along the first line ou drew. gain hang the plumb bob from the nail and trace along the string. The center of gravit is located halfwa through the thickness of the plwood under the intersection of the two lines ou drew. *Q.8 In cases (a) and (c) the center of gravit is above the base b one-half the height of the can. So (b) is the answer. In this case the center of gravit is above the base b onl a bit more than one-quarter of the height of the can ch_p3-336.indd 3 /8/7 8::33 PM

2 3 Chapter *Q.9 nswer (b). The skscraper is about 3 m tall. The gravitational field (acceleration) is weaker at the top b about 9 parts in ten million, b on the order of 4 times. The top half of the uniform building is lighter than the bottom half b about ( )( 4 ) times. Relative to the center of mass at the geometric center, this effect moves the center of gravit down, b about ( )( 4 )(5 m) mm. Q. She can be correct. If the dog stands on a relativel thick scale, the dog s legs on the ground might support more of its weight than its legs on the scale. She can check for and if necessar correct for this error b having the dog stand like a bridge with two legs on the scale and two on a book of equal thickness a phsics tetbook is a good choice. *Q. nswer (b). Visualize the hatchet as like a balanced plaground 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. Q. The free bod diagram demonstrates that it is necessar 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. *Q.3 nswer (g). In the problems we stud, the forces applied to the object lie in a plane, and the ais we choose is a line perpendicular to this plane, so it appears as a point on the free-bod diagram. It can be chosen anwhere. The algebra of solving for unknown forces is generall easier if we choose the ais where some unknown forces are acting. *Q.4 (i) nswer (b). The etension is directl proportional to the original dimension, according to F YΔL L i. (ii) nswer (e). Doubling the diameter quadruples the area to make the etension four times smaller. Q.5 Shear deformation. FIG. Q ch_p3-336.indd 3 /3/7 7:5:4 PM

3 Static Equilibrium and Elasticit 33 SOLUTIONS TO PROBLEMS Section. The Conditions for Equilibrium P. Take torques about P. τ p n + d m g d m b gd m g We want to find for which n. + mg+ mg b d mg mg + m + mb d m m m g m b g m g d m P m O CG n O n P FIG. P. P. Use distances, angles, and forces as shown. The conditions of equilibrium are: F F F + R F g F F R l F τ F cosθ Fg F cosθ sinθ R R F g O FIG. P. Section. More on the Center of Gravit P.3 The coordinates of the center of gravit of piece are. cm and 9. cm The coordinates for piece are 8. cm and. cm 8. cm 4. cm The area of each piece is 7. cm and 3. cm nd the mass of each piece is proportional to the area. Thus, and CG CG m i i m m i i m i i ( 7 cm )( cm)+ ( 3 cm )... ( 8. cm) 7. cm + 3. cm ( 7 cm )( 9 cm)+ ( 3 cm )... (. cm) 4 cm. cm FIG. P cm 685. cm 4. cm 3794 ch_p3-336.indd 33 /3/7 7:5:5 PM

4 34 Chapter P.4 The hole we can count as negative mass Call σ the mass of each unit of pizza area. CG CG CG σπ R σπ R R/ 8 R 3/4 6 m m m m P.5 Let the fourth mass (8. kg) be placed at (, ), then R R σπ ( ) R σπ ( ) m ( 3. )( 4. ) + 4 CG. + m m 4 Similarl, CG ( 3. )( 4. ) m P.6 Let σ represent the mass-per-face area. vertical strip at position, with width d and height ( 3). has mass. m 9 σ ( 3. ) d dm 9 ( 3.) /9 The total mass is 3. σ ( 3) d M dm 9 M 3. σ ( 6 + 9) d 9 d FIG. P.6 3. m M 9 σ σ The -coordinate of the center of gravit is dm CG d M ( ) 3 σ ( σ ) σ σ d m m 3794 ch_p3-336.indd 34 /8/7 8:: PM

5 Static Equilibrium and Elasticit 35 P.7 In a uniform gravitational field, the center of mass and center of gravit of an object coincide. Thus, the center of gravit of the triangle is located at 667. m, 33. m (see the Eample on the center of mass of a triangle in Chapter 9). The coordinates of the center of gravit of the three-object sstem are then: + m i i 6. kg 55. m 3. kg ( m)+ ( 5. kg) ( 3. 5 m) CG mi ( ) kg kg m CG 54. m and 4. kg m i i ( 6 CG. kg)( 7. m)+ ( 3. kg)( 33. m)+ ( 5. kg) ( m) mi 4. kg kg m CG 475. m 4. kg Section.3 Eamples of Rigid Objects in Static Equilibrium P.8 (a) For rotational equilibrium of the lowest rod about its point of support, τ. +. g g 3 cm mg4 cm m 9. g (b) (c) For the middle rod, + m cm. g + 9. g 5cm m 5. 5 g For the top rod, 55. g+. g+ 9. g 4cm m 6cm m g 3 P.9 τ mg( 3r) Tr 3r T Mgsin 45. Mg sin kg g sin 45. T ( 53)( 9. 8) N 53 m T g 77 kg 3g 3g 5 kg θ 45 m P. (a) Taking moments about P, ( Rsin 3. ) + ( Rcos 3. )( 5. cm) ( 5 N )( 3. cm) R 39. N. 4 kn The force eerted b the hammer on the nail is equal in magnitude and opposite in direction: 4. kn at 6 upward and to the right. (b) f Rsin 3. 5 N 37 N n Rcos 3. 9 N F ( 37 N) ˆi + ( 9 N) ĵ surface FIG. P ch_p3-336.indd 35 /3/7 7:5:7 PM

6 36 Chapter P. (a) F f nw F ng 8 N 5 N Taking torques about an ais at the foot of the ladder, ( 8 N)( 4. m) sin 3. + ( 5 N)( 7. 5 m) sin3. ( 5. cm) cos 3. n w Solving the torque equation, + n w [ 4. m 8 N 7. 5 m 5 N ] tan N 5. m Net substitute this value into the F equation to find Solving the equation f n w 68 N in the positive direction. F, f n w 5 N n g 8 N FIG. P. n g 3 N in the positive direction. (b) In this case, the torque equationτ gives: ( 9. m)( 8 N) sin 3. + ( 7. 5 m)( 5 N) sin3. ( 5. m) sin 6. or n w 4 N Since f 4 N and f fma μ n g, we find n w f 4 N μ 3 N ma. n g 34 n w P. (a) F f nw () F ng mg mg () τ mg L θ mg θ nl w θ cos cos + sin From the torque equation, n m g + L mg w cotθ Then, from equation (): f n m g L mg w + cotθ and from equation (): n m + g m g (b) If the ladder is on the verge of slipping when d, f n w m g n g θ m g FIG. P. then f d μ n g m / + m d/ L cotθ m + m 3794 ch_p3-336.indd 36 /3/7 7:5:8 PM

7 Static Equilibrium and Elasticit 37 P.3 Torque about the front wheel is zero.. m mg 3. m F r Thus, the force at each rear wheel is F r. mg. 94 kn The force at each front wheel is then F f mg Fr 44. kn FIG. P.3 *P.4 (a) The gravitational force on the floodlight is ( kg)(9.8 m s ) 96 N We consider the torques acting on the beam, about an ais perpendicular to the page and through the left end of the horizontal beam. H V d T N τ +( T sin 3. ) d ( 96 N) d giving T 39 N. FIG. P.4 (b) From F, H T cos 3., or H ( 39 N) cos N to the right. (c) From F,V + Tsin N, orv 96 N ( 39 N) sin 3.. (d) From the same free-bod diagram with the ais chosen at the right-hand end, we write τ H() Vd + T() + 96 N() so V (e) From F,V + Tsin N, or T + 96 N/s in N. (f) From F, H T cos 3., or H ( 39 N) cos N to the right. (g) The two solutions agree precisel. The are equall accurate. The are essentiall equall simple. But note that man students would make a mistake on the negative (clockwise) sign for the torque of the upward force V in the equation in part (d). Taking together the equations we have written, we appear to have four equations but we cannot determine four unknowns. Onl three of the equations are independent, so we can determine onl three unknowns. P.5 (a) Vertical forces on one half of the chain: T e sin 4.. N T e 9. 9 N (b) Horizontal forces on one half of the chain: T cos 4. T T m. N e m 3794 ch_p3-336.indd 37 /3/7 7:5: PM

8 38 Chapter P.6 Relative to the hinge end of the bridge, the cable is attached horizontall out a distance ( 5. m) cos. 47. m and verticall down a distance ( 5. m ) sin. 7. m. The cable then makes the following angle with the horizontal: (. +. 7) m θ tan m (a) Take torques about the hinge end of the bridge: R+ R 9. 6 kn( 4. m ) cos. T cos 7. (. 7 m)+ T sin 7. ( 4. 7 m) 98. kn( 7. m) cos. which ields T kn R T R. 4. m 5. m 7. m 9.6 kn FIG. P kn (b) F R T cos 7. or R ( kn) cos kn ( right) (c) F R 9. 6 kn + T sin kn Thus, R 9. 4 kn ( kn) sin kn 49. kn down P.7 (a) We model the horse as a particle. The drawbridge will fall out from under the horse. (b) θ α mg cos 3g cosθ m ( ms ) cos. 8 (. m) Iω mgh m ω mg sinθ 3 Solving, 3g 398. ms ω ( sinθ ) 8. m 73. rad s ( sin ). 56 rad s R R θ mg FIG. P.7(a) continued on net page 3794 ch_p3-336.indd 38 /3/7 7:5: PM

9 Static Equilibrium and Elasticit 39 (c) The linear acceleration of the center of mass of the bridge is a α m. rad s. m s The force at the hinge plus the gravitational force produce the acceleration a m s at right angles to the bridge. R ma ( kg)( m s ) cos kn R mg ma R R θ a mg FIG. P.7(c) Solving, R m g a kg m s ( + ) ( ) ( ms) sin kn Thus (d) R R 47. ˆ i+ 66. ˆj kn a ω ( 56. rad s) ( 4. m) 967. m s a R mg ma R ( kg) ( 98. m s m s ) 38.9kN mg R R Thus: R 38. 9ĵ kn FIG. P.7(d) P.8 Call the required force F, with components F F cos 5. and F Fsin 5., transmitted to the center of the wheel b the handles. Just as the wheel leaves the ground, the ground eerts no force on it. R a distances b 8. cm F F 4 N n n a forces b F : F cos. n 5 () FIG. P.8 F : Fsin. + n 5 4 N () Take torques about its contact point with the brick. The needed distances are seen to be: b R 8. cm (. 8. ) cm. cm a R b 6. cm (a) τ : Fb + Fa +( 4 N ) a, or F[ (. cm) cos 5. + ( 6. cm) sin 5. ]+ ( 4 N)( 6. cm) so 6 4 N cm F 7.45 cm 859 N continued on net page 3794 ch_p3-336.indd 39 /3/7 7:5:3 PM

10 3 Chapter (b) Then, using Equations () and (), n ( 859 N) cos N and n 4 N+( 859 N) sin 5. 6 N n n + n n n θ tan tan ( 7. 49) to the left and upward 4. kn P.9 When min, the rod is on the verge of slipping, so f ( f ) μ n 5n s ma s. From F, n T cos 37, or n. 799T. Thus, f. 5(. 799T). 399T From F, f + T sin 37 F g, n f. m F g Fg FIG. P.9 T 37. m or. 399T. 6T F g, giving T. F g Using τ for an ais perpendicular to the page and through the left end of the beam gives F F + ( F g g m g) min. sin m, which reduces to min 8m. P. Consider forces and torques on the beam. F : Rcosθ T cos 53 F : Rsinθ + T sin 53 8 N τ : ( T sin 53 ) 8 m ( 6 N) ( N) 4 m (a) (b) 6 N + 8 N m Then T 8msin 53 ( Nm) + 5 N s increases from m, this epression grows larger. From substituting back, Rcos θ [ ] cos 53 Rsin θ 8 N [ ]sin 53 Dividing, sin tanθ R θ tan cosθ + 8 N 53 R ( ) cos 53 tanθ tan s increases the fraction decreases and θ decreases. continued on net page 3794 ch_p3-336.indd 3 /3/7 7:5:5 PM

11 Static Equilibrium and Elasticit 3 (c) To find R we can work out R cos θ+ R sin θ R. From the epressions above for R cosθ and Rsinθ, R T cos 53 + T sin 53 6 NT sin 53 +( 8 N) R T 6T sin R ( ) 78( )+ 64 R t this gives R 74 N. t m, R 58 N. t 8m, R 537 N. Over the range of possible values for, the negative term dominates the positive term 8 89, and R decreases as increases. P. To find U, measure distances and forces from point. Then, balancing torques, (. 75) U 9. 4(. 5) U 88. N To find D, measure distances and forces from point B. Then, balancing torques, ( 75. ) D ( 5. )( 94. ) D N lso, notice that U D+ F g, so F. Section.4 Elastic Properties of Solids P. The definition of Y stress means that Y is the slope of the graph: strain P.3 F L Y Δ L i 6 3 Nm Y. 3 FLi ( 9. 8)( 4. ) ΔL Y P.4 (a) F F stress πr d F ( stress) π 8.5 F ( 5. Nm) π F kn (b) YΔL stress Y ( strain). 49 ( ). mm L i m Nm Li ΔL 8 stress (. 5 Nm)(. 5 m) 5. mm Y 5. Nm 3794 ch_p3-336.indd 3 /3/7 7:5:6 PM

12 3 Chapter P.5 From the defining equation for the shear modulus, we find Δ as 3 hf ( 5. m)(. N) Δ S 3. Nm 4 4. m or Δ 38. mm P.6 Count the wires. If the are wrapped together so that all support nearl equal stress, the number should be. kn. kn Since cross-sectional area is proportional to diameter squared, the diameter of the cable will be ( mm) ~ cm 5 m P.7 (a) F ( stress) (b) ( ) 3 8 π 5. m 4. N m N The area over which the shear occurs is equal to the circumference of the hole times its thickness. Thus, 3 3 ( m) ( πr) t π 5. m 5. So, m ( ) 4 8 F Stress. 57 m 4. N m N t 3. ft F FIG. P.7(b) P.8 The force acting on the hammer changes its momentum according to m vf vi mvi + F( Δ t) mvf so F Δt Hence, 3. kg. ms. ms F N. s B Newton s third law, this is also the magnitude of the average force eerted on the spike b the hammer during the blow. Thus, the stress in the spike is: 3 F 88. N 7 stress 97. Nm π.3 m / 4 stress 97. and the strain is: strain Y. 7 N m Nm ch_p3-336.indd 3 /3/7 7:5:7 PM

13 Static Equilibrium and Elasticit 33 P.9 Consider recompressing the ice, which has a volume. 9V. ΔV ΔP B ( 9. Nm )(. 9 ) N m 9. V i P.3 Let the 3. kg mass be mass #, with the 5. kg mass, mass #. ppling Newton s second law to each mass gives: ma T mg () and ma mg T () where T is the tension in the wire. T Solving equation () for the acceleration gives: a m g and substituting this into equation () ields: m m T m g m g T Solving for the tension T gives mmg T 3. kg 5. kg 9. 8 m s m + m 8. kg N FL From the definition of Young s modulus, Y i, the elongation of the wire is: ( Δ L ) ΔL TL i ( N)(. m) 9 3 Y. Nm 3. mm π. m P.3 Part of the load force etends the cable and part compresses the column b the same distance Δ : Y Δ Y s sδ F + s F Δ Y / + Y s s/ s 8 5 N 7 π / π. 7 / m ΔP ΔPVi P.3 B ΔV / Vi ΔV 8 3 ΔPVi ( 3. Nm) m (a) ΔV m 3 B. Nm (b) The quantit of water with mass. 3 3 kg occupies volume at the bottom m m. 946 m. So its densit is 3 3. kg 3 9. kg m m (c) With onl a 5% volume change in this etreme case, liquid water is indeed nearl incompressible ch_p3-336.indd 33 /3/7 7:5:8 PM

14 34 Chapter dditional Problems P.33 Let n and n B be the normal forces at the points of support. Choosing the origin at point with F and τ, we find: 4 4 n + n 8. g 3. g and B ( ) ( ) g g 5. n B ( 5. ) 5. m 5. m FIG. P.33 B The equations combine to give n N and b B N. *P.34 Model the stove as a uniform 68 kg bo. Its center of mass is at its geometric center, N 4 inches behind its feet at the front corners. F g ssume that the light oven door opens to be horizontal and that a person stands on its outer end, inches in front n of the front feet. FIG. P.34 We find the weight F g of a person standing on the oven door with the stove balanced on its front feet in equilibrium: τ + ( 68 kg 9. 8 m s 4 in. n F g in. ) F g 58 N If the weight of the person is greater than this, the stove can tip forward. This weight corresponds to mass 5.8 kg, so the person could be a child. If the oven door is heav (compared to the backsplash) or if the front feet are significantl far behind the front corners, the maimum weight will be significantl less than 58 N. P.35 With as large as possible, n and n will both be large. The equalit sign in f μ s n will be true, but the less-than sign will appl in f < μ s n. Take torques about the lower end of the pole. Setting f n cosθ+ F g cosθ f sinθ. 576n, the torque equation becomes n (. 576 tanθ ) + F g n θ f Fg θ f FIG. P.35 n d Since n >, it is necessar that. 576 tanθ < tanθ > θ > 6. d. sinθ < 78ft sin ft 3794 ch_p3-336.indd 34 /4/7 5:7:3 PM

15 Static Equilibrium and Elasticit 35 P.36 When the concrete has cured and the pre-stressing tension has been released, the rod presses in on the concrete and with equal force, T, the concrete produces tension in the rod. (a) In the concrete: stress 8 6 ΔL. N m Y strain Y Thus, Li ΔL 6 stress ( 8. Nm)( 5. m) Y 3. 9 Nm L i or 4 ΔL 4. m. 4 mm T 6 (b) In the concrete: stress 8. N m, so c ( ) 6 4 T 8. Nm 5. m 4. kn (c) (d) (e) For the rod: T L Δ L Y R ΔL i steel so ΔL TLi Y R steel 4 ( 4. N)( 5. m) 4 5. m. Nm. m. mm 3 ( ) The rod in the finished concrete is. mm longer than its unstretched length. To remove stress from the concrete, one must stretch the rod.4 mm farther, b a total of 4. mm. For the stretched rod around which the concrete is poured: T L Δ Y total L T L Δ L total steel R i i 3 4. T.5 m P.37 (a) See the diagram. or Y R m 4 ( 5. m )(. N m ) 48. kn steel R T (b) If. m, then τ O ( 7 N)(. m) ( N)( 3. m) ( 8. N)( 6. m) + ( T sin 6. )( 6. m) O R 6. N 7 N 3. m 3. m 8. N Solving for the tension gives: T 343 N. FIG. P.37 From F, R T cos 6. 7 N. From F, R N T 98 sin N. (c) If T 9 N: τ O ( 7 N) ( N)( 3. m) ( 8. N) 6. m + ( 9 N) sin 6. ( 6. m) Solving for gives: 53. m ch_p3-336.indd 35 /3/7 7:5: PM

16 36 Chapter *P.38 The 39 N is the weight of the uniform gate, which is 3 m wide. The hinges are.8 m apart. The eert horizontal forces and C. Onl one hinge eerts a vertical force. We assume it is the upper hinge. (a) Free bod diagram: Statement: uniform 4.-kg farm gate, 3. m wide and.8 m high, supports a 5.-N bucket of grain hanging from its latch as shown. The gate is supported b hinges at two corners. Find the force each hinge eerts on the gate. C B 3. m 5 N 39 N.8 m (b) From the torque equation, 738 N m C 4 N.8 m Then 4 N. lso B 44 N. The upper hinge eerts 4 N to the left and 44 N up. The lower hinge eerts 4 N to the right. FIG. P.38 P.39 Using F F τ, choosing the origin at the left end of the beam, we have (neglecting the weight of the beam) F R Tcosθ F R + Tsinθ F g and τ Fg ( L+ d) + T sinθ( L+ d). Solving these equations, we find: Fg ( L+ d) (a) T sinθ ( L+ d) (b) R F ( L+ d) cotθ g L+ d R FL g L+ d FIG. P.39 P.4 τ point gives 3 3 ( T cos 5. ) sin 65. Tsin cos ( N)( cos 65. ) + ( N) cos 65. From which, T 465 N. 46 kn l T cos 5. 3l 4 N T sin 5. N From F, H T cos N( toward right). 33 kn From F, V 3 N T sin N( upward). 58 kn H V 65. FIG. P ch_p3-336.indd 36 /3/7 7:5: PM

17 Static Equilibrium and Elasticit 37 P.4 We interpret the problem to mean that the support at point B is frictionless. Then the support eerts a force in the direction and F F B F F F B ( 3 + ) g and τ ( 3 g )(. ) ( g )( 6. ) + F (. ) B These equations combine to give F F 647. B 5 N FIG. P.4 F 7. *P.4 Choosing torques about the hip joint, τ gives L + L T L 35 N sin. N 3 From which, T. 7kN. 5 N Let R compression force along spine, and from R T T cos.. 65 kn F FIG. P.4 (c) You should lift with our knees rather than with our back. In this situation, with a load weighing onl N, ou can make the compressional force in our spine about ten times smaller b bending our knees and lifting with our back as straight as possible. P.43 From the free-bod diagram, the angle that the string tension makes with the rod is θ and the perpendicular component of the string tension is T sin 8.. Summing torques around the base of the rod gives T τ : ( 4. m)( N) cos 6 + T ( 4. m) sin8 N cos 6. sin 8. F : F H T cos. T N F V 6 N F H T cos N F : F + V T sin. N F H FIG. P.43 and F N T sin N V 3794 ch_p3-336.indd 37 /3/7 7:5:4 PM

18 38 Chapter P.44 (a) Just three forces act on the rod: forces perpendicular to the sides of the trough at and B, and its weight. The lines of action of and B will intersect at a point above the rod. The will have no torque about this point. The rod s weight will cause a torque about the point of intersection as in Figure.5(a), and the rod will not be in equilibrium unless the center of the rod lies verticall below the intersection point, as in Figure.5(b). ll three forces must be concurrent. Then the line of action of the weight is a diagonal of the rectangle formed b the trough and the normal forces, and the rod s center of gravit is verticall above the bottom of the trough. F g O FIG. P.44(a) B (b) In Figure (b), O cos 3. BO cos 6. and O BO O O L + + O L + cos3 cos6 L 3 6 cos. cos. θ 3. O F g 6. B FIG. P.44(b) So O cosθ L and θ 6. P.45 (a) Locate the origin at the bottom left corner of the cabinet and let distance between the resultant normal force and the front of the cabinet. Then we have F cos 37. μ n () F + sin 37. n 4 () τ n(. 6 ) 4(. 3) + sin 37. (. 6) cos 37. (. 4) (3) From (), n 4 sin N From (3), (. 6) cm to the left of the front edge (b) cos 37. From (), μ k In this case, locate the origin at the bottom right corner of the cabinet. Since the cabinet is about to tip, we can use τ to find h: τ 4(. 3) ( 3 cos 37. ) h h 3 cos 37. FIG. P m 3794 ch_p3-336.indd 38 /3/7 7:5:9 PM

19 Static Equilibrium and Elasticit 39 P.46 (a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet. τ F (. m) + ( 4 N)(. 3 m) ielding F ( 4 N )(. 3 m ) N. m F f + N, or f N Thus, F 4 N + n, so n 4 N μ s f N n 4 N. 3 (c) ppl F at the upper rear corner and directed so θ+ φ 9. to obtain the largest possible lever arm. θ. m tan m Thus, φ Sum the torques about the lower front corner of the cabinet: F (. m) + (. 6 m) + ( 4 N)(. 3 m) so N m F 3 N.7 m Therefore, the minimum force required to tip the cabinet is 3 N applied at 3. above the horizontal at the upper left corner. P.47 (a) We can use F F and τ with pivot point at the contact on the floor. Then F T μ n s F n Mg mg, and τ Mg ( L cosθ) + mg L cosθ θ T ( Lsin ) Solving the above equations gives m μs sinθ cosθ f M cosθ μs sinθ FIG. P.47 This answer is the maimum value for M if μs < cot θ. If μ θ s cot, the mass M can increase without limit. It has no maimum value. P Mg mg 4 N 4 N T F' φ θ.3 m n.6 m FIG. P.46 L/ θ θ f F. m L/ n. m f n (b) t the floor, we have the normal force in the -direction and frictional force in the -direction. The reaction force then is R n +( μ n) ( M + m) g + μ t point P, the force of the beam on the rope is s F T +( Mg) g M + μ ( M + m) s s 3794 ch_p3-336.indd 39 /3/7 7:5:3 PM

20 33 Chapter *P.48 Suppose that a bar eerts on a pin a force not along the length of the bar. Then, the pin eerts on the bar a force with a component perpendicular to the bar. The onl other force on the bar is the pin force on the other end. For F, this force must also have a component perpendicular to the bar. Then each of the forces produces a torque about the center of the bar in the same sense. The total torque on the bar is not zero. The contradiction proves that the bar can onl eert forces along its length. FIG. P.48 *P.49 (a) The height of pin B is N (. m) sin m B The length of bar BC is then. m 5. m n BC 77. m sin 45. Consider the entire truss: FIG. P.49(a) F n N + n Which gives n C 634 N. Then, C τ ( N). cos 3. + n C [. cos cos 45. ] n N n 366 N C C n C (b) Joint : C B F : C B sin N so C B 73 N T C n 366 N F : CBcos. + TC 3 T C ( 73 N) cos N Joint B: F : ( 73 N ) cos 3. C cos 45. BC C BC ( 73 N ) cos N cos 45. N B C B 73 N C BC FIG. P.49(b) 3794 ch_p3-336.indd 33 /3/7 7:5:34 PM

21 Static Equilibrium and Elasticit 33 *P.5 Considering the torques about the point at the bottom of the bracket ields: W(. 5 m) F(. 6 m) so F.833W (a) With W 8. N, F.833(8 N) N. (b) Differentiate with respect to time: df dt.833 dw dt The force eerted b the screw is increasing at the rate df dt.833(.5 N s).5 N s P.5 From geometr, observe that cosθ 4 and θ For the left half of the ladder, we have F T R () F R + n 686 N () T T τ top 686N(. cos 755. ) + T (. sin 755. ) FIG. P.5 n ( 4. cos 755. ) (3) For the right half of the ladder we have F R T F nb R (4) τ top nb ( 4. cos 755. ) T(. sin 755. ) (5) Solving equations through 5 simultaneousl ields: (a) T 33 N (b) n 49 N and n B 57 N (c) R 33 N and R 57 N The force eerted b the left half of the ladder on the right half is to the right and downward ch_p3-336.indd 33 /3/7 7:5:36 PM

22 33 Chapter P.5 Imagine graduall 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. s P reaches its maimum value, the clinder will turn clockwise microscopicall to stress the welds at both contact points and make both forces of friction increase to their maimum values. comparison: To make a four-legged table start to slide across the floor, ou must push on it hard enough to counterbalance the maimum static friction forces on all four legs together. FIG. P.5 When it is on the verge of slipping, the clinder is in equilibrium. F : f n s n F : P+ n + f F g τ : P f + f μ and f μ n s P grows so do f and f. s Therefore, since μ s, f n and f n then P+ n + F g 4 n n 4 () and P n n 3 + n So P+ n F g becomes P+ 4 P F 4 g 4 3 or () 8 3 P F g Therefore, P F g 3 8 P.53 (a) F k( Δ L), Young s modulus is Y F/ FL L L ( i Δ / ΔL ) i Thus, Y kl and k Y i L i ΔL ΔL ΔL Y ( Δ (b) W Fd ( k) d d L ) L Y i L i 3794 ch_p3-336.indd 33 /3/7 7:5:37 PM

23 Static Equilibrium and Elasticit 333 P.54 (a) Take both balls together. Their weight is 3.33 N and their CG is at their contact point. F : + P3 P P F : + τ : PR+ PR ( R+ Rcos 45. ) + Substituting, P 333. N P 333. N N + P ( R+ Rcos 45. ) PR +( 333. N) R ( 333. N ) R( + cos 45. ) + PR ( + cos 45. ) ( 333. N) cos 45. P cos 45. P 67. N so P 67. N 3 P N P FIG. P.54(a) (b) Take the upper ball. The lines of action of its weight, of P, and of the normal force n eerted b the lower ball all go through its center, so for rotational equilibrium there can be no frictional force. F : ncos 45. P 67. N n cos N n cos N n sin 45. FIG. P.54(b) P F : nsin N gives the same result. P.55 F : + 38 N F g + 3 N F g 7 N Take torques about her feet: τ : 38 N (. m ) + ( 7 N ) +( 3 N) FIG. P m P.56 The tension in this cable is not uniform, so this becomes a fairl difficult problem. dl F L Y t an point in the cable, F is the weight of cable below that point. Thus, F µ g where µ is the mass per unit length of the cable. Then, Li Li dl L d g Y d gli µ µ Y (. 4)( 9. 8)( 5) ( ) m cm 3794 ch_p3-336.indd 333 /8/7 8::5 PM

24 334 Chapter P.57 (a) F m Δv Δ t (.. ). kg. s ms 4 5 N F 4 5 N (b) stress Nm (. m )(. m ) (c) Yes. This is more than sufficient to break the board. P.58 Let θ represent the angle of the wire with the vertical. The radius of the circle of motion is r (. 85 m) sinθ. For the mass: v Fr mar m mrω r [ ] T sin θ m (. 85 m) sinθ ω Further, T Y ( strain ) or T Y ( strain) r T FIG. P.58 mg Thus, Y ( strain) m(. 85 m) ω, giving ω or ω 573. rad s Y ( strain) π 39. m 7. Nm. m. 85 m (. kg)(. 85 m) 4 3 ( ) P.59 (a) If the acceleration is a, we have P ma and P + n Fg. Taking the origin at the center of gravit, the torque equation gives P ( L d) + Ph nd Solving these equations, we find d HL CG h P P Fg L d ah g n F g FIG. P.59 (b) ah ( If P, then d g. ms )( 5. m ) 98. ms. 36 m. (c) Using the given data, P 36 N and P 553 N. Thus, P 36ˆi+ 553ˆj N ch_p3-336.indd 334 /3/7 7:5:4 PM

25 Static Equilibrium and Elasticit 335 P.6 When the car is on the point of rolling over, the normal force on its inside wheels is zero. F ma: n mg F ma: f m v R Take torque about the center of mass: fh n d. mg f d mg h v ma Then b substitution m R h mgd v gdr ma h FIG. P.6 wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover. NSWERS TO EVEN PROBLEMS P. F + R Fg ; F R ; F cosθ F g cosθ F sinθ P.4 see the solution P.6.75 m P.8 (a) 9. g (b) 5.5 g (c) 49. g P. (a).4 kn at 6 upward and to the right. (b) F surface ( 37 N) ˆi+ ( 9 N) ˆj mg mg P. (a) f + cotθ ; ng m m g L ( + ) ( m / + md / L) cotθ (b) μ m + m P.4 (a) 39 N (b) 339 N to the right (c) (d) (e) 39 N (f ) 339 N to the right (g) Both are equall accurate and essentiall equall simple. We appear to have four equations, but onl three are independent. We can determine three unknowns. P.6 (a) 35.5 kn (b).5 kn to the right (c) 4.9 kn down P.8 (a) 859 N (b) 4 kn at 36.9 above the horizontal to the left P. (a) see the solution (b) θ decreases (c) R decreases P.. N m P.4 (a) 73.6 kn (b).5 mm P.6 ~cm P P mm P.3 (a).53 8 m 3 (b).9 3 kg m 3 (c) With onl a 5% change in volume in this etreme case, liquid water is indeed nearl incompressible in biological and student-laborator situations ch_p3-336.indd 335 /3/7 7:5:4 PM

26 336 Chapter P.34 The weight must be 58 N or more. The person could be a child. We assume the stove is a uniform bo with feet at its corners. We ignore the masses of the backsplash and the oven door. If the oven door is heav, the minimum weight for the person might be somewhat less than 58 N. P.36 (a).4 mm (b) 4. kn (c). mm (d).4 mm (e) 48. kn P.38 (a) See the solution. The weight of the uniform gate is 39 N. It is 3. m wide. The hinges are separated verticall b.8 m. The bucket of grain weighs 5. N. One of the hinges, which we suppose is the upper one, supports the whole weight of the gate. Find the components of the forces that both hinges eert on the gate. (b) The upper hinge eerts 4 N to the left and B 44 N up. The lower hinge eerts C 4 N to the right. kn P kn; 33. ˆ i+ 58. ˆj P.4 (a).7 kn (b).65 kn (c) You should lift with our knees rather than with our back. In this situation, ou can make the compressional force in our spine about ten times smaller b bending our knees and lifting with our back as straight as possible. P.44 (a) see the solution (b) 6. P.46 (a) N (b).3 (c) 3 N at 3. above the horizontal to the right P.48 ssume a strut eerts on a pin a force with a component perpendicular to the length of the strut. Then the pin must eert a force on the strut with a perpendicular component of this size. For translational equilibrium, the pin at the other end of the strut must also eert the same size force on the strut in the opposite direction. Then the strut will feel two torques about its center in the same sense. It will not be in equilibrium, but will start to rotate. The contradiction proves that we were wrong to assume the eistence of the perpendicular force. The strut can eert on the pin onl a force parallel to its length. P.5 (a) 66.7 N (b) increasing at.5 N s P.5 If either static friction force were at less than its maimum value, the clinder would rotate b a microscopic amount to put more stress on some welds and to bring that friction force to its maimum value. P 3F g 8 P.54 (a) P.67 N; P 333. N; P N (b).36 N P cm P rad s P.6 See the solution. wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover ch_p3-336.indd 336 /3/7 7:5:43 PM