EVALUATE: If the angle 40 is replaces by (cable B is vertical), then T = mg and

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Corey Sutton
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1 58 IDENTIY: ppl Newton s 1st law to the wrecing ball Each cable eerts a force on the ball, directed along the cable SET UP: The force diagram for the wrecing ball is setched in igure 58 (a) T cos 40 mg = 0 mg (4090 g)(980 m/s ) T = = = cos40 cos40 (b) T sin 40 T = 0 T 4 = Tsin 40 = N N igure 58 EVLUTE: If the angle 40 is replaces b (cable is ertical), then T = mg and T = IDENTIY: ppl to each bloc a = 0 SET UP: Tae + perpendicular to the incline and + parallel to the incline The free-bod diagrams for each bloc, and, are gien in igure 514 (a) or, gies T1 wsinα = 0 and T1 = wsinα (b) or bloc, gies T1 T wsinα = 0 and T = wsinα (c) for each bloc gies n = n = wcosα (d) or α 0, T1 = T 0 and n = n w or α 90, T1 = w, T = w and n = n = 0 EVLUTE: The two tensions are different but the two normal forces are the same igure 514a, b 51IDENTIY: ppl to each bloc Each bloc has the same magnitude of acceleration a

2 SET UP: ssume the pulle is to the right of the 400 g bloc There is no friction force on the 400 g bloc, the onl force on it is the tension in the rope The 400 g bloc therefore accelerates to the right and the suspended bloc accelerates downward Let + be to the right for the 400 g bloc, so for it a = a, and let be downward for the suspended bloc, so for it a = a + (a) The free-bod diagrams for each bloc are gien in igures 51a and b (b) applied to the 400 g bloc gies T = (400 g) aand T 100 N 50 m/s a = = = 400 g 400 g (c) applied to the suspended bloc gies mg T and m = T 100 N 137 g g = a 980 m/s 50 m/s = (d) The weight of the hanging bloc is mg = (137 g)(980 m/s ) = 134 N This is greater than the tension in the rope; T = 075mg EVLUTE: Since the hanging bloc accelerates downward, the net force on this bloc must be downward and the weight of the hanging bloc must be greater than the tension in the rope Note that the blocs accelerate no matter how small m is It is not necessar to hae m > 400 g, and in fact in this problem m is less than 400 g igure 51a, b ma to the composite object of eleator plus student ( m tot = 850 g ) and also to the student ( w = 550 N ) The eleator and the student hae the same acceleration SET UP: Let + be upward The free-bod diagrams for the composite object and for the student are gien in igure 54a and b T is the tension in the cable and n is the scale reading, the normal force the scale eerts on the student The mass of the student is m= w/ g = 561 g (a) applied to the student gies n mg n mg 450 N 550 N a = = = 178 m/s The eleator has a downward acceleration of 178 m/s m 561 g 670 N 550 N (b) 14 m/s a = = 561 g (c) n = 0 means a = g The student should worr; the eleator is in free-fall (d) applied to the composite object gies T mtotg = mtota T = m ( tot a + g) In part (a), T = (850 g)( 178 m/s m/s ) = 680 N In part (c), a = g and T = 0 EVLUTE: In part (b), T = (850 g)(14 m/s m/s ) = 10,150 N The weight of the 54IDENTIY: ppl = composite object is 8330 N When the acceleration is upward the tension is greater than the weight and when the acceleration is downward the tension is less than the weight

3 igure 54a, b 59(a) IDENTIY: Constant speed implies a = 0 ppl Newton s 1st law to the bo The friction force is directed opposite to the motion of the bo SET UP: Consider the free-bod diagram for the bo, gien in igure 59a Let be the horizontal force applied b the worer The friction is inetic friction since the bo is sliding along the surface igure 59a n mg = 0 n= mg So f = μ n= μ mg f = 0 = f = μ mg = (00)(11 g)(980 m/s ) = N (b) IDENTIY: Now the onl horizontal force on the bo is the inetic friction force ppl Newton s nd law to the bo to calculate its acceleration Once we hae the acceleration, we can find the distance using a constant acceleration equation The friction force is f = μ mg, just as in part (a) SET UP: The free-bod diagram is setched in igure 59b igure 59b f μmg = μ g = (00)(980 m/s ) = 196 m/s a Use the constant acceleration equations to find the distance the bo traels: = 0, 0 = 350 m/s, a = 196 m/s, 0 =? = 0 + a( 0) 0 0 (350 m/s) 0 = = = 31 m a ( 196 m/s ) EVLUTE: The normal force is the component of force eerted b a surface perpendicular to the surface Its magnitude is determined b In this case n and mg are the onl ertical forces and a = 0, so n= mg lso note that f and n are proportional in magnitude but perpendicular in direction

4 545IDENTIY: ppl to each bloc SET UP: or bloc use coordinates parallel and perpendicular to the incline Since the are connected b ropes, blocs and also moe with constant speed (a) The free-bod diagrams are setched in igure 545 (b) The blocs moe with constant speed, so there is no net force on bloc ; the tension in the rope connecting and must be equal to the frictional force on bloc, μ = (035) (50 N) = 9 N (c) The weight of bloc C will be the tension in the rope connecting and C; this is found b considering the forces on bloc The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the component of the weight along the ramp, the friction on bloc and the tension in the second rope Thus, the weight of bloc C is w = 9 N + w (sin369 + μ cos369 ) = 9 N + (50 N)(sin (035)cos 369 ) = 310 N C The intermediate calculation of the first tension ma be aoided to obtain the answer in terms of the common weight w of blocs and, wc = w ( μ + (sinθ+ μcos θ) ), giing the same result (d) ppling Newton s Second Law to the remaining masses ( and C) gies: a = g( wc μwcosθ wsin θ) ( w + wc) = 154m s EVLUTE: efore the rope between and is cut the net eternal force on the sstem is zero When the rope is cut the friction force on is remoed from the sstem and there is a net force on the sstem of blocs and C igure IDENTIY: We can use the analsis done in Eample 53 s in that eample, we assume friction is negligible SET UP: rom Eample 53, the baning angle β is gien b tan β = lso, n= mg/ cos β gr 650 mi/h = 91 m/s (91 m/s) (a) tan β = and β = 10 The epression for tan β does not inole (980 m/s )(5 m) the mass of the ehicle, so the truc and car should trael at the same speed (115 g)(980 m/s ) 4 4 (b) or the car, n car = = N and ntruc = ncar = N, since cos10 m = m truc car EVLUTE: The ertical component of the normal force must equal the weight of the ehicle, so the normal force is proportional to m 557IDENTIY: ppl to the motion of the pilot The pilot moes in a ertical circle The apparent weight is the normal force eerted on him t each point a rad is directed toward the center of the circular path (a) SET UP: the pilot feels weightless means that the ertical normal force n eerted on the pilot b the chair on which the pilot sits is zero The force diagram for the pilot at the top of the path is gien in igure 557a

5 igure 557a mg rad g = R Thus = gr = (980 m/s )(150 m) = 3834 m/s 1 m 3600 s = (3834 m/s) 138 m/h 3 = 10 m 1 h (b) SET UP: The force diagram for the pilot at the bottom of the path is gien in igure 557b Note that the ertical normal force eerted on the pilot b the chair on which the pilot sits is now upward igure 557b w = 700 N, so m = w = 7143 g g 3 1 h 10 m = (80 m/h) = 7778 m/s 3600 s 1 m (7778 m/s) Thus n = 700 N g = 3580 N 150 m > g n mg = m R n= mg + m R This normal force is the pilot s apparent weight EVLUTE: In part (b), n m since the acceleration is upward The pilot feels he is much heaier than when at rest The speed is not constant, but it is still true that a at each point of the motion = R rad / 568 IDENTIY: ppl to the brush Constant speed means a = 0 Target ariables are two of the forces on the brush SET UP: Note that the normal force eerted b the wall is horizontal, since it is perpendicular to the wall The inetic friction force eerted b the wall is parallel to the wall and opposes the motion, so it is erticall downward The free-bod diagram is gien in igure 568 n cos531 = 0 n= cos531 f = μn= μcos531 igure 568 sin531 w f = 0 sin 531 w μ cos531 = 0

6 (sin 531 μ cos531 ) = w w = sin 531 μ cos531 w 10 N (a) = = 169 N sin 531 μ cos531 sin 531 (015) cos531 = (b) n= cos531 = (169 N)cos531 = 101 N EVLUTE: In the absence of friction w= sin531, which agrees with our epression 57IDENTIY: ppl = ma to the passenger to find the maimum allowed acceleration Then use a constant acceleration equation to find the maimum speed SET UP: The free-bod diagram for the passenger is gien in igure 57 gies n mg n= 16mg, so a = 060 g = 588 m s 0 0 = 30 m, a = 588 m s, = 0 so = + a ( ) gies = 50 m s 0 0 EVLUTE: larger final speed would require a larger alue of a, which would mean a larger normal force on the person igure IDENTIY: ppl to each bloc orces between the blocs are related b Newton s 3rd law The target ariable is the force loc is pulled to the left at constant speed, so bloc moes to the right at constant speed and a = 0 for each bloc SET UP: The free-bod diagram for bloc is gien in igure 583a n is the normal force that eerts on f = μ n is the inetic friction force that eerts on loc moes to the right relatie to, and f opposes this motion, so f is to the left Note also that acts just on, not on T f = 0 T = f = 040 N igure 583a n w = 0 n = 140 N f = μ n = (030)(140 N) = 040 N

7 SET UP: The free-bod diagram for bloc is gien in igure 583b igure 583b n is the normal force that bloc eerts on bloc Newton s third law n and n are equal in magnitude and opposite in direction, so n = 140 N f is the inetic friction force that eerts on loc moes to the left relatie to and f opposes this motion, so f is to the right f = μ n = (030)(140 N) = 040 N n and f are the normal and friction force eerted b the floor on bloc ; f = μ n Note that bloc moes to the left relatie to the floor and f opposes this motion, so f is to the right n w n = 0 n= w + n = 40 N N = 560 N Then f = μ n= (030)(560 N) = 168 N f + T + f = 0 = T + f + f = 040 N N N = 5 N EVLUTE: Note that f and f are a third law action-reaction pair, so the must be equal in magnitude and opposite in direction and this is indeed what our calculation gies 593 IDENTIY: ppl to the bloc and to the plan SET UP: oth objects hae a = 0 n be the normal force between the plan and the bloc and Let be the normal force between the bloc and the incline Then, n = wcosθ and n = n + 3wcosθ = 4wcos θ The net frictional force on the bloc is μ ( n + n ) = μ 5wcosθ To moe at constant speed, this must balance the component of the bloc s weight along the incline, so 3wsinθ = μ 5wcos θ, and μ = tanθ = tan 37 = EVLUTE: In the absence of the plan the bloc slides down at constant speed when the slope angle and coefficient of friction are related b tanθ = μ or θ = 369, μ = 075 smaller μ is needed when the plan is present because the plan proides an additional friction force 5117 IDENTIY: The elocit is tangent to the path The acceleration has a tangential component when the speed is changing and a radial component when the path is curing SET UP: a rad is toward the center of curature of the path a tan is parallel to when the speed is increasing and antiparallel to when the speed is decreasing The net force is proportional to a The diagram is setched in igure 5117 n

8 igure 5117 EVLUTE:, a, and all change during the motion

Note: Referred equations are from your textbook.

Note: Referred equations are from your textboo 50 IDENTIFY: pply Newton s first law to the car SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp EXECUTE: (a) The free-body