L04hw - Dynamics Homework Solutions 5 Ed 6 Ed 5 Ed 6 Ed Q 1...

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Unformatted text preview: Dynamics Homework Solutions 5 Ed. 6 Ed. 5 Ed. 6 Ed. Q: 1, x, 3, 8; P: 1, 6, 8, x, 14, 16, 23, x Q: 1, 2, 3, 7; P: 1, 5, 7, 11, 13, 15, 19, 23 Q: 15, 21; P: 27, 81, 36*, 39, x, 40, x, 44, 47, 49, 60*, 61* Q: 17, 19; P: 27, 31, 34*, 37, 41, 43, 45, 47, 49, 51, 61*, 63* P1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.25 m/s Given: Solution: kg m/s kg m/s N P5. A 20-kg box rests on a table (a) What is the weight of the box and the normal force acting on it? (b) A 10-kg box is placed on top of the 20-kg box, as shown on the diagram. Determine the normal force that the table exerts on the 20.0-kg box and the normal force that the 20.0-kg box exerts on the 10.0-kg box. N12 Nt1 WE2 N21 Nt1 W E1 WE1 N1t N1t (a) kg (b) m/s N 1 Top box: kg m/s =98.0 N Bottom box: kg m/s kg m/s N P7. What average force is needed to accelerate a 7.00-gram pellet from rest to 125 m/s over a distance of 0.800 m along the barrel of a rifle? Given: g m/s m/s m ? kg Find: Solution: m/s m m/s kg m/s m/s N P11. A particular race car can cover a quarter-mile track m in 6.40 s starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and race car is 485 kg, what horizontal force must the road exert on the tires? Given: m/s s m kg ? ? Find: Solution: 2 m m m/s g's 2 s kg m/s N N P13. An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680g. What are the maximum and minimum forces the motor should exert on the supporting cable? Given: Solution: F a a kg m/s F W W 1. If the acceleration is up: kg m/s kg m/s N 2. If the acceleration is down: kg m/s kg m/s N P15. A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of the person's regular weight. Calculate the acceleration of the elevator, and find the direction of acceleration. Solution: 3 N mg Two forces are acting on the person: Gravity and Normal force. The person is moving together with the elevator The normal force is given to be: Second Law: m/s m/s and points downwards The acceleration is equal to P19. A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end. Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 N (b) 60.0 N (c) 90.0 N Solution: Simplifications: 1. The pulley is massless 2. The cord is massless Therefore: 1. The tension along the rope will always be the same. The acceleration of the weight and the box is the same in magnitude and opposite in direction Both objects are not moving Weight: Box: ! " ! acceleration is zero. ! " " 4 (a) " N N N N N N kg kg N N N' ' '' N #$% #&% Note: N N cannot be negative P23. Arlene is to walk across a "high wire" strung horizontally between two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0. If her mass is 50.0 kg, what is the tension in the rope at this point? Solution: 80. 0 T mg T 10.0 In equilibrium, Arlene does not have any acceleration Second Law along the vertical direction gives us: ! cos () kg m/s N ! cos cos N Q: 17, 19; P: 27, 31, 34*, 37, 41, 43, 45, 47, 49, 51, 61, 63* P27. Two snowcats tow a housing unit to a new location at McMurdo Base, Antarctica. The sum of the forces * and + exerted on the unit by the horizontal cables is parallel to the line L, and * N . Determine + and the magnitude of * + Solution: 5 y L 50 FA FAx F Ax 30 F B FB Y FBx x If the total force is along the line L, then the x-components of the two forces must be equal and opposite in sign, so that they will cancel out. * + + + * sin + sin * + sin * + * cos + cos sin * sin N * sin The total force is along L, so all we have to do is add the y-components of the two vectors and that will give us the total force. * + * cos + cos N P31. A block with mass m is on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block with mass m (a) Draw a free-body diagram for each block (b) Determine formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and the rope. Solution: (a) FN m1 T T Fg1 m2 Fg2 6 (b) ! ! P34.* The two masses shown on the diagram are each initially 1.80 m above the ground, and the massless frictionless pulley is 4.8 m above the ground. What maximum height does the lighter object reach after the system is released? T a T g a 4. 8 m m 1g FN m1 v1 4.8 m m1 m2 1. 80 m m 1g m2 g 3.60 m m2 m2 g 1. The masses experience tension from the rope and are moving in opposite directions: Mass 1: ! Mass 2: ! ! ! ! m/s For mass I, the motion is with constant acceleration: m/s m/s When mass 2 hits the ground, it will have fallen by 1.80 m, therefore mass 1 will have climbed up by 1.80 m , and thus it will be hanging 3.60 m above the ground. m ? what will be the velocity of mass 1 at that moment? 7 m/s m/s m/s m From this moment on, there will be no tension in the rope and mass 1 will be free falling. m/s m/s ? the max. height m/s velocity is zero at max. height 2 m/s 1 falling. m m/s m/s distance traveled from the position when the mass 1 began free Thus the max. position above the ground will be: m P37. A force of 48.0 N is required to start a 5.0 kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor? (b) If the 48-N force continues, the box accelerates at 0.70 m/s . What is the coefficient of kinetic friction? Solution: FN F=48.0 N Ff r mg (a) N , , kg N m/s - , (b) , , . , N . . kg kg . N m/s m/s N 8 P41. A 15.0-kg box is released on a 32 incline and accelerates down the incline at 0.30 m/s . Find the friction force impeding its motion. What is the coefficient of kinetic friction? Solution: m Fgx Fgy 32 a Fg 32 , , kg , kg m/s sin . cos kg m/s N The box is moving . cos , . N m/s cos P43. (a) A box sits at rest on a rough 30 inclined plane. Draw the free-body diagram, showing all the forces acting on the box. (b) How would the diagram change if the box were sliding down the plane? (c) How would it change if the box were sliding up the plane after an initial shove? Solution: m Ffrs Fgy 30 FN Fgx a Ff rk Fgy m FN Fgx a a m FN Fgx Ff rk 30 30 30 Fgy 30 30 Fg Fg Fg (a) Static friction, up (b) Kinetic friction, up (c) Kinetic friction, down P45. The coefficient of kinetic friction for a 22-kg bobsled on a track is 0.10 . What force is required to push it down a 6.0 incline and achieve a speed of 60 km/h at the end of 75 m? Solution: 9 F Ffrk Fgy 6 FN Fgx a 6 Fg In order to start from rest and achieve a velocity m km/h m/s s after distance m, the acceleration must be equal to: m/s m/s m cos kg m/s cos The acceleration is then caused by the net force along the incline: . , . . , Friction is: N kg N , sin m/s N kg m/s sin P47. A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 4.0 m/s ? Solution: , , . . pointing against the direction m/s m/s Note: the acceleration is negative - that is points opposite to the initial velocity m/s m/s m/s , . . m/s m/s m/s m/s m m/s 10 P49. A flatbed truck is carrying a heavy crate. The coefficient of static friction between the crate and the bed of the truck is 0.75. What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab of the truck? Solution: In order for the crate not to slide, the acceleration of the crate must be equal to that of the truck. , / / , / - - / - m/s P51. A child slides down a slide with a 28 incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child. Solution: With friction: sin sin sin . cos . . cos cos Without friction: sin sin sin 0 0 Speed with friction is half that without friction: 0 sin sin . cos . cos . sin sin P61. The 75-kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back. The static coefficient s of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that friction forces are both at a maximum. Ignore his grip on the rope. . Solution: 11 ...
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This document was uploaded on 09/21/2009.

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