PROBLEM SOLUTION

Transcription

1 PROLEM , Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 15 kn. m= 35, Mg = 35 1 kg 3 F = 15 1 N v 1 = 4 km/hr = m/s mv Ft = (35 1 kg)( m/s) (15 1 N) t = 1 t = 59.6 s t = 4min19 s PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 695

2 PROLEM Solve Problem assuming that the brakes are applied only on the wheels of car. PROLEM The subway train shown is traveling at a speed of 3 mi/h when the brakes are fully applied on the wheels of cars and C, causing them to slide on the track, but are not applied on the wheels of car. Knowing that the coefficient of kinetic friction is.35 between the wheels and the track, determine (a) the time required to bring the train to a stop, (b) the force in each coupling. Weights of cars: W = W = 8, lb, W = 1, lb Masses of cars: C = C = = m m 484 lb s /ft, m 316 lb s /ft For each car the normal force (upward) is equal in magnitude to the weight of the car. N = N = 8, lb N = 1, lb C Friction forces: F = (.35)(8, ) = 8, lb Stopping data: v1 = 3 mi/h = 44 ft/s, v =. F F C = (brakes not applied) = (a) pply the principle of impulse-momentum to the entire train. m = m + m + m = 874 lb s /ft C F = F + F + F = 8, lb C mv + Ft = mv 1 mv ( 1 v) (874)(44) t = = = s t = 1.69 s F 8, (b) Coupling force F : pply the principle of impulse-momentum to car alone. mv + Ft+ F t= 1 (484)(44) + (8,)(1.688) + F (1.688) = F = 19,39 lb F = 19,39 lb (compression) PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 78

3 PROLEM (Continued) Coupling force F C : pply the principle of impulse-momentum to car C alone. mv+ Ft F t= C 1 C C (484)(44) + () F (1.688) = C F = 86 lb F = 86 lb (compression) C C PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 79

4 PROLEM The triple jump is a track-and-field event in which an athlete gets a running start and tries to leap as far as he can with a hop, step, and jump. Shown in the figure is the initial hop of the athlete. ssuming that he approaches the takeoff line from the left with a horizontal velocity of 1 m/s, remains in contact with the ground for.18 s, and takes off at a 5 angle with a velocity of 1 m/s, determine the vertical component of the average impulsive force exerted by the ground on his foot. Give your answer in terms of the weight W of the athlete. Vertical components mv + ( P W) Δ t = mv Δ t =.18 s 1 W + ( Pv W)(.18) = (1)(sin 5 ) g (1)(sin 5 ) Pv = W + W (9.81)(.18) Pv = 6.1W PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 74

5 PROLEM One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 1 in., the height of the first bounce of the ball must be in the range 53 in. h 58 in. Determine the range of the coefficient of restitution of the tennis balls satisfying this requirement. Uniform accelerated motion: v = gh v = gh Coefficient of restitution: Height of drop v e = v h e = h h = 1 in. Height of bounce 53 in. h 58 in. Thus, e.78 e PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 75

6 PROLEM Two identical pool balls of mm diameter, may move freely on a pool table. all is at rest and ball has an initial velocity v = v i. (a) Knowing that b = 5 mm and e =.7, determine the velocity of each ball after impact. (b) Show that if e = 1, the final velocities of the balls from a right angle for all values of b. Geometry at instant of impact: b 5 sinθ = = d θ = 61.3 Directions n and t are shown in the figure. Principle of impulse and momentum: all : all : all, t-direction: mvsinθ + = m( v) t ( v) t= vsinθ (1) all, t-direction: + = mv ( ) ( v ) = () alls and, n-direction: mv cos θ + + m ( v ) + m ( v ) t t n n ( v ) + ( v ) = v cosθ (3) n n Coefficient of restitution: ( v) n ( v) n = e[ v cos θ ] (4) (a) e =.7. From Eqs. (1) and (), ( v ) =.87489v (1) t ( v ) t= () PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 77

7 PROLEM (Continued) From Eqs. (3) and (4), ( v ) + ( v ) =.4843v (3) n n ( v ) ( v ) = (.7)(.4843 v ) (4) n n Solving Eqs. (5) and (6) simultaneously, (b) e = 1. Eqs. (3) and (4) become ( v ) =.7648 v ( v ) =.41167v n n = ( ) n + ( ) t v v v v = (.7648 v ) + ( =.8779v ( v) n.7648v tan β = = =.8337 ( v) t.87489v β = ϕ = 9 θ β = = 4.1 v v =.878v =.41v ( v ) + ( v ) = v cosθ (3) n n ( v ) ( v ) = v cosθ (4) n n Solving Eqs. (3) and (4) simultaneously, ( v ) =, ( v ) = v cosθ n ut ( v) t= v sin θ, and ( v) t= t v is in the t-direction and v is in the n-direction; therefore, the velocity vectors form a right angle. PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 773

8 PROLEM Two cars of the same mass run head-on into each other at C. fter the collision, the cars skid with their brakes locked and come to a stop in the positions shown in the lower part of the figure. Knowing that the speed of car just before impact was 5 mi/h and that the coefficient of kinetic friction between the pavement and the tires of both cars is.3, determine (a) the speed of car just before impact, (b) the effective coefficient of restitution between the two cars. (a) t C: Conservation of total momentum: m = m = m 5 mi/h = ft/s mv + mv = mv + mv v = v + v (1) Work and energy. Care (after impact): U U 1 T1 = m( v ) T = (1) = μ m g(1 ft) 1 m( v ) mkmg(1) = ( v ) = ()(1 ft)(.3)(3. ft/s ) f k = ft/s v = 15.6 ft/s = F T + U = T PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 78

9 PROLEM (Continued) Car (after impact): (b) U 1 T1 = m( v ) T = 1 T + U = T 1 1 v = μ m g(3) k 1 m ( v ) m g(3) μ k = ()(3 ft)(.3)(3. ft/s ) ( v ) = ft/s v = ft/s From (1) v = v + v = Relative velocities: ( v v ) e= v v ( ) e = (7.613) e = =.8 (37.53) v = 3. ft/s =.6 mi/h e =.3 PROPRIETRY MTERIL. 13 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual may be displayed, 783