F f v 1 = c100(10 3 ) m h da 1h 3600 s b =

Transcription

1 The 2-Mg car has a velocity of v 1 = 100km>h when the v km/h driver sees an obstacle in front of the car. It takes 0.75 s for him to react and lock the brakes, causing the car to skid. If the car stops when it has traveled a distance of 175 m, determine the coefficient of kinetic friction between the tires and the road. Free-ody Diagram: The normal reaction N on the car can be determined by writing the equation of motion along the y axis and referring to the free-body diagram of the car, Fig. a, + c F y = ma y ; N (9.81) = 2000(0) N = N Since the car skids, the frictional force acting on the car can be computed from F f = m k N = m k (19 620). Principle of Work and Energy: y referring to Fig. a, notice that only which is negative. The initial speed of the car is m>s. Here, the skidding distance of the car is s. T 1 + U 1-2 = T (2000)( ) + C -m k (19 620)s D = 0 s = m k does work, The distance traveled by the car during the reaction time is s =v 1 t = 27.78(0.75) = m. Thus, the total tal distance traveled by the car before it stops is s = s +s 175 = m k F f v 1 = c100(10 3 ) m h da 1h 3600 s b = This work uriprotected s by United States copyright laws sale of any part of his work (including on the World Wide Web) m k = 0.255

2 The two blocks and have weights W = 60 lb and W = 10 lb. If the kinetic coefficient of friction between the incline and block is m k = 0.2, determine the speed of after it moves 3 ft down the plane starting from rest. Neglect the mass of the cord and pulleys. Kinematics: The speed of the block and can be related by using position coordinate equation s + (s - s ) = l 2s - s = l 2 s - s = 0 s = 2 s = 2(3) = 6ft 2v - v = 0 (1) Equation of Motion: pplying Eq. 13 7, we have + F y = ma y ; N - 60a 4 5 b = (0) N = 48.0 lb Principle of Work and Energy: y considering the whole system, W which acts in the direction of the displacement does positive work. W and the friction force F f = m k N = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite po direction to that of displacement Here, W is being displaced vertically (downward) d) 3 and W is being displaced vertically (upward) s. Since blocks are 5 s and at rest initially, T 1 = 0. pplying Eq. 14 7, we have 0 + W 3 5 s - F f s - W s = 1 v 2 m m 2 v work (3)R (3) - 10(6) = v v2 2 32provided 2 1 and Eqs. (1) and (2) yields T 1 + a U 1-2 = T = 60v v v 2 2 protected 2 (2) This w is by United States copyrigh laws 2and is ro solely for the use of instructors in teaching their courses assessing student learning. Dissemination or sale of any part of o this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. v = 3.52 ft>s v = ft s

3 The man at the window wishes to throw the 30-kg sack on the ground. To do this he allows it to swing from rest at to point C, when he releases the cord at u = 30. Determine the speed at which it strikes the ground and the distance R. 8m 8m 16 m u C T + U - C = T C (9.81)8 cos 30 = 1 2 (30)v2 C y C = m>s R D T + U - D = T D (9.81)(16) = 1 2 (30) v2 D v D = 17.7 m>s During free flight: (+T) s = s 0 + v 0 t a ct 2 16 = 8 cos sin 30 t (9.81)t2 t t = 0 Solving for the positive root: t = s ( : + ) s = s 0 + v 0 t s = 8 sin cos 30 (2.0784) s = m Thus, This work is protected by United States copyright laws sale of any part of this work (including on the World Wide Web) R = = 33.0 m lso, (v D ) x = cos 30 = m>s (+T)(v D ) x = sin (2.0784) = m>s n D = 2(10.097) 2 + (14.559) 2 = 7.7 m>s

4 lb block rests on the smooth semicylindrical surface. n elastic cord having a stiffness k = 2lb>ft is attached to the block at and to the base of the semicylinder at point C.If the block is released from rest at ( u = 0 ), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant u = 45. Neglect the size of the block. C k 2lb/ft u 1.5 ft +b F n = ma n ; 2 sin 45 = a v2 1.5 b v = ft>s T 1 + U 1-2 = T (2)Cp(1.5) - l 0D (2)c 3p 2 4 (1.5) - l 0 d - 2(1.5 sin 45 ) = 1 2 a b(5.844)2 l 0 = 2.77 ft This work is protected by United States copyright lawsws sale of any part of this work (including on the World Wide Web)

5 * The 500-kg elevator starts from rest and travels upward with a constant acceleration a c = 2m>s 2. Determine the power output of the motor M when t = 3s. Neglect the mass of the pulleys and cable. M + c F y = ma y ; 3T - 500(9.81) = 500(2) T = N E 3s E - s P = l 3 v E = v P When t = 3s, (+ c) v 0 + a c t v E = 0 + 2(3) = 6m>s v P = 3(6) = 18 m>s P O = (18) P O = 35.4 kw This work is protected by United States copyright laws sale of any part of this work (including on theworld Wide Web)