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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term 2012 Exam 1: Practice Problems Solutions One-Dimensional Kinematics: vx (t) ! vx,0 = ! ! ! dr ! dv v= , a= dt dt t " =t # t " =0 ax (t " ) dt ", x(t) ! x0 = Initial conditions: x0 and vx, 0 are the values at t = 0 . Newton’s Second Law: Force, Mass, Acceleration: Newton’s Third Law: © Peter Dourmashkin 2012 ! ! F1,2 = !F2,1 ! ! F = ma t " =t # t " =0 vx (t " ) dt " Problem 1: Elevator A person of mass m p stands on a scale in an elevator of mass m p . The scale reads the magnitude of the force F exerted on it from above in a downward direction. Starting at rest at t = 0 the elevator moves upward, coming to rest again at time t = t 0 . The downward acceleration of gravity is g . The acceleration of the elevator during this period is shown graphically above and is given analytically by a y (t) = ! " 2! t. t0 (1) a) Find the maximum speed of the elevator. b) Find the total distance traveled by the elevator. Solutions: Part a) The velocity is the integral of the acceleration. Inspection of the graph shows the integral increases until t = t 0 = 2 where the velocity reaches its maximum value. That value is given by the area of the triangle above the axis. v y,max = (1 / 2)! (t0 / 2) = ! t . 4 0 (2) Part b) t t$ 2# ' # v y (t) ! v y0 = " a y (t)dt = " & # ! t ) dt = # t ! t 2 . t0 ( t0 0 0% (3) Because v y0 = 0 , the y-component of the velocity is v y (t) = ! t " ! 2 t . t0 (4) Integrating again to find the position t t$ # ' # # 3 y(t) ! y0 = " v y (t)dt = " & # t ! t 2 ) dt = t 2 ! t . t0 ( 2 3t0 0 0% (5) Because we choose y0 = 0 , y(t) = ! 2 ! 3 t " t . 2 3t0 The total distance traveled by the elevator is the position at time t = t 0 y(t0 ) = ! 2 ! 3 ! 2 t " t = t0 . 2 0 3t0 0 6 (6) Problem 2: Traffic Violation A car is driving through a green light at t = 0 located at x = 0 with an initial speed vc,0 = 12 m ! s-1 . The acceleration of the car as a function of time is given by #%0; 0 < t < t1 = 1 s ac = $ . -3 &%!(6 m " s )(t ! t1 ); 1 s < t < t2 a) Find the speed and position of the car as a function of time. b) Graph the speed and position of the car as a function of time. c) A bicycle rider is riding at a constant speed of vb,0 and at t = 0 is 17 m behind the car. The bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle. Solution: a) We need to integrate the acceleration for both intervals. The first interval is easy, the speed is constant. For the second integral we need to be careful about the endpoints of the integral and the fact that the integral is the change in speed so we must subtract vc (t1 ) = vc0 $vc0 ; 0 < t < t1 = 1 s & t . vc (t) = % -3 v (t ) + !(6 m " s )(t ! t ) ; 1 s < t < t c 1 1 2 # & t1 ' After integrating we get #vc0 ; 0 < t < t1 = 1 s % . vc (t) = $ t -3 2 %&vc0 ! (3 m " s )(t ! t1 ) t1 ; 1 s < t < t2 Now substitute the endpoint so the integral to finally yield #%vc0 = 12 m ! s-1; 0 < t < t1 = 1 s . vc (t) = $ -1 -3 2 %&12 m ! s " (3 m ! s )(t " t1 ) ; 1 s < t < t2 For this one dimensional motion the change in position is the integral of the speed so t1 $ & xc (0) + " (12 m ! s-1 )dt; 0 < t < t1 = 1 s & 0 xc (t) = % . t & x (t ) + 12 m ! s-1 # (3m ! s-3 )(t # t )2 dt; 1 s < t < t 1 2 & c 1 " t1 ' ( ) Upon integration we have # xc (0) + (12 m ! s-1 )t; 0 < t < t1 = 1 s % xc (t) = $ -1 -3 3 % xc (t1 ) + (12 m ! s )(t " t1 ) " (1 m ! s )(t " t1 ) & ( We choose our coordinate system ) t t1 ; 1 s < t < t2 such . that xc (0) = 0 , therefore xc (t1 ) = (12 m ! s )(1 s)=12 m . So after substituting in the endpoints of the integration interval we have that -1 #%(12 m ! s-1 )t; 0 < t < t1 = 1 s . xc (t) = $ -1 -3 3 12 m+(12 m ! s )(t " t ) " (1 m ! s )(t " t ) ; 1 s < t < t 1 1 2 &% b) Graph the speed and position of the car as a function of time. Solution: The graphs of the speed and position are shown below. c) A bicycle rider is riding at a constant speed of vb,0 and at t = 0 is 17 m behind the car. The bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle. Solution: we are looking for the instant that t2 the car has come to rest. So we use our expression for the speed for the interval 1 s < t < t2 , 0 = vc (t2 ) = 12 m ! s-1 " (3 m ! s-3 )(t2 " t1 )2 . We can solve this for t2 : (t2 ! t1 )2 = 4 s 2 . We have two solutions: (t2 ! t1 ) = 2 s or (t2 ! t1 ) = !2 s . The t2 = t1 ! 2 s = 1 s ! 2 s = ! 1 s does not apply to our time interval and so second solution t2 = t1 + 2 s = 1 s + 2 s = 3 s . During the position of the car at t2 is then given by xc (t2 ) = 12 m+(12 m ! s-1 )(t2 " t1 ) " (1 m ! s-3 )(t2 " t1 )3 = 12 m+(12 m ! s-1 )(2 s) " (1 m ! s-3 )(2 s)3 = 28 m . Since the bicycle is traveling at a constant speed with an initial position xb0 = !17 m , the position of the bicycle is given by xb (t) = !17 m + vbt . The bicycle and car intersect at instant t2 = 3 s : xb (t2 ) = xc (t2 ) . Therefore !17 m + vb (3 s) = 28 m . So the speed of the bicycle is vb = (28 m + 17 m) = 15 m ! s-1 . (3 s) Problem 3 Catching a bus At the instant a traffic light turns green, a car starts from rest with a given constant acceleration, 5.0 ! 10"1 m # s-2 . Just as the light turns green, a bus, traveling with a given constant speed, 1.6 ! 101 m " s-1 , passes the car. The car speeds up and passes the bus some time later. How far down the road has the car traveled, when the car passes the bus? Solution: I. Understand – get a conceptual grasp of the problem Think about the problem. How many objects are involved in this problem? Two, the bus and the car. How many different stages of motion are there for each object? Each object has one stage of motion. For each object, how many independent directions are needed to describe the motion of that object? One, only. What information can you infer from the problem? The acceleration of the car, the velocity of the bus, and that the position of the car and the bus are identical when the bus just passes the car. Sketch qualitatively the position of the car and bus as a function of time (Figure 4.12). Figure 4.12 Position vs. time of the car and bus. What choice of coordinate system best suits the problem? Cartesian coordinates with a choice of coordinate system in which the car and bus begin at the origin and travel along the positive x-axis (Figure 4.13). Draw arrows for the position coordinate function for the car and bus. Figure 4.13 A coordinate system for car and bus. II. Devise a Plan - set up a procedure to obtain the desired solution Write down the complete set of equations for the position and velocity functions. There are two objects, the car and the bus. Choose a coordinate system with the origin at the traffic light with the car and bus traveling in the positive x-direction. Call the position function of the car, x1 (t) , and the position function for the bus, x2 (t) . In general the position and velocity functions of the car are given by 1 x1 (t) = x1,0 + vx10t + ax,1t 2 2 vx,1 (t) = vx10 + ax,1t In this example, using both the information from the problem and our choice of coordinate system, the initial position and initial velocity of the car are both zero, x1,0 = 0 and vx10 = 0 , and the acceleration of the car is non-zero ax,1 ! 0 . So the position and velocity of the car is given by 1 a t2 , 2 x,1 vx,1 (t) = ax,1t . x1 (t) = The initial position of the bus is zero, x2,0 = 0 , the initial velocity of the bus is non-zero, x2 (ta ) = vx,20ta vx,20 ! 0 , and the acceleration of the bus is zero, ax,2 = 0 . So the position function for the bus is x2 (t) = vx,20t . The velocity is constant, vx,2 (t) = vx,20 . Identify any specified quantities. The problem states: “The car speeds up and passes the bus some time later.” What analytic condition best expresses this condition? Let t = ta correspond to the time that the car passes the bus. Then at that time, the position functions of the bus and car are equal, x1 (ta ) = x2 (ta ) . How many quantities need to be specified in order to find a solution? There are three independent equations at time t = ta : the equations for position and velocity of the car x1 (ta ) = 1 ax,1ta2 2 vx,1 (ta ) = ax,1ta , and the equation for the position of the bus, . There is one ‘constraint condition’ x1 (ta ) = x2 (ta ) . The six quantities that are as yet unspecified are x1 (ta ) , x2 (ta ) , vx,1 (ta ) , vx,20 , ax , ta So you need to be given at least two numerical values in order to completely specify all the quantities; for example the distance where the car and bus meet. The problem specifying the initial velocity of the bus, vx,20 , and the acceleration, ax , of the car with given values. III. Carry our your plan – solve the problem! The number of independent equations is equal to the number of unknowns so you can design a strategy for solving the system of equations for the distance the car has traveled in terms of the velocity of the bus vx,20 and the acceleration of the car ax,1 , when the car passes the bus. Let’s use the constraint condition to solve for the time t = ta where the car and bus meet. Then we can use either of the position functions to find out where this occurs. Thus the constraint condition, x1 (ta ) = x2 (ta ) becomes 1 ax,1ta2 = vx,20ta . 2 We can solve for this time, ta = 2 vx,20 ax,1 . Therefore the position of the car at the meeting point is x1 (ta ) = ! v $ v2 1 1 ax,1ta2 = ax,1 # 2 x,20 & = 2 x,20 . 2 2 ax,1 " ax,1 % IV. Look Back – check your solution and method of solution Check your algebra. Do your units agree? The units look good since in the answer the two sides agree in units, ! m 2 % s-2 # !" m #$ = & -2 ' " m% s $ and the algebra checks. Substitute in numbers. Suppose ax = 5.0 ! 10"1 m # s-2 and vx,20 = 1.6 ! 101 m " s-1 , Introduce your numerical values for vx,20 and ax , and solve numerically for the distance the car has traveled when the bus just passes the car. Then ta = 2 vx,20 ax,1 (1.6 ! 10 m " s ) = 6.4 ! 10 s =2 (5.0 ! 10 m " s ) 1 -1 1 #1 ( () ( -2 ) 2 2 1.6 ! 101 m " s-1 vx,20 x1 (ta ) = 2 = 2 = 1.0 ! 103 m #1 -2 ax,1 5.0 ! 10 m " s ) Check your results. Once you have an answer, think about whether it agrees with your estimate of what it should be. Some very careless errors can be caught at this point. Is it possible that when the car just passes the bus, the car and bus have the same velocity? Then there would be an additional constraint condition at time t = ta , that the velocities are equal, vx,1 (ta ) = vx,20 . Thus vx,1 (ta ) = ax,1ta = vx,20 implies that ta = vx,20 ax,1 . From our other result for the time of intersection ta = 2vx,20 ax,1 . But these two results contradict each other, so it is not possible. Problem 4 Softball Catch A softball is hit over a third baseman’s head. The third baseman, as soon as the ball is hit, turns around and runs straight backwards at a constant speed of 7.0 m ! s "1 for a time interval 2.0s and catches the ball at the same height it left the bat. The third baseman was initially 18 m from home plate. What is the initial speed and angle of the softball when it left the bat? Ignore air resistance. Solution: It’s probably best to choose the origin at the point where the ball leaves the bat at time t = 0 . This way, both the initial and final vertical components of the ball’s position are zero. As suggested by the diagram, use v0 for the ball’s initial speed, and ! 0 for the angle of the ball’s initial velocity with respect to the horizontal. Use the horizontal direction, from the batter to the fielder, as the +iˆ direction, and the upward vertical as the +jˆ direction. The equation for the horizontal component of the position of the ball is given by xb (t) = vx,0t (1) The third baseman is running at a constant velocity vt ,0 = v1 . Note: We are using “ t ” in the subscript to denote the third baseman and “ b ” in the subscript to denote the ball. The equation for the position of the third baseman is xt (t) = xt ,0 + vt ,0t (2) So at the end of the time interval tcatch = !t is the third baseman is a horizontal distance xt (tcatch ) = xt ,0 + vt ,0tcatch = xt ,0 + v1 !t = 1.8 " 101 m + (7.0 m # s $1 )(2.0s) = 32 m (3) from the point where the ball hits the bat. At the instant tcatch the third baseman catches the ball, we have that Eq. (1) becomes xb (tcatch ) = xt (tcatch ) (4) Setting t = tcatch into Eq. (1) and Eq. (2) and then substituting into the left hand side and right hand side of Eq. (4) yields vx,0tcatch = xt ,0 + vt ,0tcatch (5) Solve Eq. (5) for the horizontal component of the initial velocity of the ball: vx0 = xt ,0 t + vt ,0 = 18m / 2s + 7.0 m ! s "1 = 16 m ! s "1 (6) The equation for the vertical component of the velocity is given by v y (t ) = v y 0 ! g t (7) v y (ttop ) = 0 (8) 0 = v y 0 ! gttop (9) At the top of the fight, Substituting Eq. (8) into Eq. (7) yields The ball reached its highest point at a time ttop = !t / 2 = 1.0s after it was hit. We use this value for the instant the ball reaches the highest point of its trajectory and solve Eq. (9) for the initial vertical component of the velocity that is given by v y 0 = gttop = g ("t / 2 ) = (9.8 m # s !2 )(1.0s) = 9.8 m # s !1 . The initial speed and angle are then (10) v0 = vx20 + v y20 = (16 m # s ) + (9.8 m # s ) "1 2 "1 2 $v % $ 9.8 m # s "1 % ! 0 = tan "1 & y 0 ' tan "1 & = 31.5!. "1 ' ( 16 m # s ) ( vx 0 ) = 18.8 m # s "1 (11) Review: The range that the ball traveled and the initial angle and speed of the ball are consistent with reasonable values that you may encounter either playing or watching a softball game. Problem 5: Jumping off a cliff A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the shore. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at an acceleration that varies in time according to ax = b1t and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , b1 , h , and the gravitational constant g . You may neglect all air resistance. Solution: While the person is running, since the acceleration varies in time, in order to find the x-component of the velocity, we integrate the x-component of the acceleration with respect to time t t 0 0 vx (t ) ! vx 0 = " ax (t )dt = " (b1t )dt . (1) = 12 b1t 2 Since the runner started from rest vx 0 = 0 and therefore vx (t ) = 12 b1t 2 . (2) We now integrate the x-component of the velocity with respect to time to find the displacement t t 0 0 x(t ) ! x0 = " vx (t )dt = " ( 12 b1t 2 )dt . (3) = 16 b1t 3 Choose an origin from the point the runner began, so x0 = 0 , and the position function is x(t ) = 16 b1t 3 . (4) Let t = t1 denote the instant the runner reaches the cliff, then x(t1 ) = d = 16 b1t13 . (5) Hence 1/ 3 ! 6d " t1 = # $ . % b1 & (6) The x-component of the velocity of the runner at t = t1 is then ! 6d " vx (t1 ) = b t = b # $ % b1 & 2 1 2 11 2/3 1 2 1 . (7) Let’s restart our clock at t = 0 when the runner leaves the cliff. Let’s pick our origin at the point of departure. Since the person is in free fall when in the air and we can neglect air resistance, the runner has acceleration ay = !g . So the y-component of the position of the runner is given by 1 y (t ) = y0 ! g t 2 . 2 (8) 1 y (t ) = h ! g t 2 . 2 (9) At t = 0 , y0 = 0 , so Let t = t2 denote the instant the runner hits the water, then 1 y (t2 ) = !h = ! g t2 2 . 2 (10) So we can solve for the time of flight, 1/ 2 ! 2h " t2 = # $ % g & . (11) The horizontal position of the runner from the edge of the cliff is given by x(t ) = vx 0t . (12) Since we know the runner leaves the cliff with an x-component of the velocity equal to the x-component of the velocity of the runner at t = t1 , we have that ! 6d " x(t ) = 12 b1 # $ % b1 & 2/3 t. (13) At t = t2 , the horizontal distance to the rock is then ! 6d " x(t2 ) = s = b # $ % b1 & 1 2 1 2/3 ! 6d " t2 = b # $ % b1 & 1 2 1 2/3 1/ 2 ! 2h " # $ % g & . (14) or ! 6d " s= b # $ % b1 & 1 2 1 2/3 1/ 2 ! 2h " # $ . % g & (15) Problem 6 Accelerating Wedge ! A 45o wedge is pushed along a table with constant acceleration A (see figure for choice of axes and positive directions) according to an observer at rest with respect to the table. A block of mass m slides without friction down the wedge. Find its acceleration with respect to an observer at rest with respect to the table. Write down a plan for finding the magnitude of the acceleration of the block. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Also clearly state any assumptions you make. Be sure you include any free body force diagrams or sketches that you plan to use. Solution: Choose a coordinate system for the block and wedge as shown in the figure below. Then ! A = Ax,w î where Ax,w is the x-component of the acceleration of the wedge. We shall apply Newton’s Second Law to the block sliding down the wedge. Since the wedge is accelerating, there is a constraint relation between the x - and y - components of the acceleration of the block. In order to find that constraint we choose a coordinate system for the wedge and block sliding down the wedge shown in the figure below. We shall find the constraint relationship between the components of the accelerations of the block and wedge by a geometric argument. From the figure above, we see that tan ! = Therefore yb . l " (xb " xw ) (1) yb = (l ! (xb ! xw )) tan " . (2) If we differentiate Eq. (2) twice with respect to time noting that d 2l =0 dt 2 (3) we have that d 2 yb dt 2 " d 2x d 2 xw % = ! $ 2b ! ' tan ( . dt 2 & # dt (4) Therefore ab, y = !(ab,x ! Ax,w ) tan " (5) recalling that Ax,w = d 2 xw dt 2 . (6) We now draw a free body force diagram for the block (figure below) Newton’s Second Law in the î - direction becomes N sin ! = mab,x . (7) N cos ! " mg = mab, y (8) and the ĵ -direction becomes We can solve for the normal force from Eq. (7): N= mab,x sin ! We now substitute Eq. (5) and Eq. (9) into Eq. (8) yielding (9) mab,x sin ! cos ! " mg = m("(ab,x " Aw,x ) tan ! ) . (10) We now clean this up yielding mab,x (cotan ! + tan ! ) = m(g + Aw,x tan ! ) (11) Thus the x-component of the acceleration is then ab,x = g + Aw,x tan ! (12) cotan ! + tan ! From Eq. (5), the y -component of the acceleration is then # g + Aw,x tan " & ab, y = !(ab,x ! Aw,x ) tan " = ! % ! Aw,x ( tan " . $ cotan " + tan " ' (13) This simplifies to ab, y = Aw,x ! g tan " (14) cotan " + tan " When ! = 45! , cotan 45! = tan 45! = 1 , and so Eq. (12) becomes ab,x = g + Aw,x (15) 2 and Eq. (14) becomes ab, y = A! g . 2 (16) The magnitude of the acceleration is then 2 ! g + Aw,x $ ! Aw,x ' g $ a= a +a = # & +# & 2 % " 2 % " 2 b,a 2 b, y 2 ! g 2 + Aw,x $ a= # &. 2 % " 2 (17) Problem 7 Pulleys and Ropes Constraint Conditions Consider the arrangement of pulleys and blocks shown in the figure. The pulleys are assumed massless and frictionless and the connecting strings are massless and unstretchable. Denote the respective masses of the blocks as m1 , m2 and m3 . The upper pulley in the figure is free to rotate but its center of mass does not move. Both pulleys have the same radius R . a) How are the accelerations of the objects related? b) Draw force diagrams on each moving object. c) Solve for the accelerations of the objects and the tensions in the ropes. Solution: Choose an origin at the center of the upper pulley. Introduce coordinate functions for the three moving blocks, y1 , y2 and y3 . Introduce a coordinate function yP for the moving pulley (the pulley on the lower right in the figure). Choose downward for positive direction; the coordinate system is shown in the figure below then. a) The length of string A is given by l A = y1 + yP + ! R (1) where ! R is the arc length of the rope that is in contact with the pulley. This length is constant, and so the second derivative with respect to time is zero, 0= d 2l A d 2 y1 d 2 yP = 2 + 2 = a y ,1 + a y , P . dt 2 dt dt (2) Thus block 1 and the moving pulley’s components of acceleration are equal in magnitude but opposite in sign, a y , P = !a y ,1 . (3) lB = ( y3 " yP ) + ( y2 " yP ) + ! R = y3 + y2 " 2 yP + ! R (4) The length of string B is given by where ! R is the arc length of the rope that is in contact with the pulley. This length is also constant so the second derivative with respect to time is zero, d 2lB d 2 y2 d 2 y3 d 2 yP 0 = 2 = 2 + 2 ! 2 2 = a y ,2 + a y ,3 ! 2a y , P . dt dt dt dt (5) We can substitute Equation (3) for the pulley acceleration into Equation (5) yielding the constraint relation between the components of the acceleration of the three blocks, 0 = a y ,2 + a y ,3 + 2a y ,1 . (6) b) Free Body Force diagrams: ! ! The forces acting on block 1 are: the gravitational force m1g and the pulling force T1,r of the string acting on the block 1. Since the string is assumed to be massless and the pulley is assumed to be massless and frictionless, the tension TA in the string is uniform and equal in magnitude to the pulling force of the string on the block. The free body diagram is shown below. Newton’s Second Law applied to block 1 is then ˆj : m g ! T = m a . 1 A 1 y ,1 (7) ! The forces on the block 2 are the gravitational force m2g and the string holding the ! block, T2,r , with magnitude TB . The free body diagram for the forces acting on block 2 are shown below. Newton’s second Law applied to block 2 is ˆj : m g ! T = m a . 2 B 2 y ,2 (8) ! The forces on the block 3 are the gravitational force m3g and the string holding the ! block, T3,r , with magnitude TB . The free body diagram for the forces acting on block 3 are shown below. Newton’s second Law applied to block 3 is ˆj : m g ! T = m a . 3 B 3 y ,3 ! ! The forces on the moving pulley are the gravitational force mP g = 0 (the pulley is ! assumed massless); string B pulls down on the pulley on each side with a force, TP ,B , ! which has magnitude TB . String A holds the pulley up with a force TP , A with the magnitude TA equal to the tension in string A . The free body diagram for the forces acting on the moving pulley are shown below. (9) Newton’s second Law applied to the pulley is ˆj : 2T ! T = m a = 0 . B A P y, P (10) Since the pulley is massless we can use this last equation to determine the condition that the tension in the two strings must satisfy, 2TB = TA (11) We are now in position to determine the accelerations of the blocks and the tension in the two strings. We record the relevant equations as a summary. 0 = a y ,2 + a y ,3 + 2a y ,1 (12) m1 g ! TA = m1 a y ,1 (13) m2 g ! TB = m2 a y ,2 (14) m3 g ! TB = m3 a y ,3 (15) 2TB = TA . (16) There are five equations with five unknowns, so we can solve this system. We shall first use Equation (16) to eliminate the tension TA in Equation (13), yielding m1 g ! 2TB = m1 a y ,1 . (17) We now solve Equations (14), (15) and (17) for the accelerations, a y ,2 = g ! TB m2 TB m3 2T a y ,1 = g ! B . m1 a y ,3 = g ! (18) (19) (20) We now substitute these results for the accelerations into the constraint equation, Equation (12), 0=g# ! 1 TB T 4T 1 4 " + g # B + 2 g # B = 4 g # TB $ + + %. m2 m3 m1 & m2 m3 m1 ' (21) We can now solve this last equation for the tension in string B , TB = 4g 4 g m1 m2 m3 . = ! 1 1 4 " m1 m3 + m1 m2 + 4 m2 m3 + + $ # % m2 m3 m1 & (22) From Equation (16), the tension in string A is TA = 2TB = 8 g m1 m2 m3 . m1 m3 + m1 m2 + 4 m2 m3 (23) We find the acceleration of block 1 from Equation (20), using Equation (22) for the tension in string B, a y ,1 = g ! 2TB 8 g m2 m3 m m + m1 m2 ! 4 m2 m3 . =g! =g 1 3 m1 m1 m3 + m1 m2 + 4 m2 m3 m1 m3 + m1 m2 + 4 m2 m3 (24) We find the acceleration of block 2 from Equation (18), using Equation (22) for the tension in string B, a y ,2 = g ! TB 4 g m1 m3 !3 m1 m3 + m1 m2 + 4 m2 m3 . =g! =g m2 m1 m3 + m1 m2 + 4 m2 m3 m1 m3 + m1 m2 + 4 m2 m3 (25) Similarly, we find the acceleration of block 3 from Equation (19), using Equation (22) for the tension in string B, a y ,3 = g ! m m ! 3 m1 m2 + 4 m2 m3 TB 4 g m1 m2 . =g! =g 1 3 m3 m1 m3 + m1 m2 + 4 m2 m3 m1 m3 + m1 m2 + 4 m2 m3 As a check on our algebra we note that 2 a1, y + a2, y + a3, y = 2g m1 m3 + m1 m2 ! 4 m2 m3 !3 m1 m3 + m1 m2 + 4 m2 m3 m m ! 3 m1 m2 + 4 m2 m3 +g +g 1 3 m1 m3 + m1 m2 + 4 m2 m3 m1 m3 + m1 m2 + 4 m2 m3 m1 m3 + m1 m2 + 4 m2 m3 = 0. (26) Problem 8 Block and Pulley A block of mass m1 rests motionless on an inclined plane that makes an angle ! with the horizontal. The coefficient of static friction between the block and plane is µs . A massless inextensible string is attached to one end of the block, passes over a fixed pulley, around a second freely suspended pulley, and is finally attached to a fixed support. The left hand part of the string is parallel to the plane. The sections of the string coming off the suspended pulley are vertical. The pulleys are massless, but an object of mass m2 is hung from the suspended pulley. Gravity acts downward. Part 1: Assume that if there were no friction the block would slide up the inclined plane. In what follows, express your answers in terms of m1 , m2 , ! , µs and the acceleration of gravity g (or some subset of these parameters). a) Draw separate free body diagrams for the system consisting of the object hanging from the suspended pulley plus the suspended pulley, and the block on the inclined plane, showing and labeling carefully, all the forces that act on the objects. b) Find the magnitude of the static friction force. c) What is the condition on the masses of the blocks such that the block on the incline plane just slides upward. Part 2: Now assume that the block on the incline plane is sliding upward. The coefficient of kinetic friction between the block and plane is µ k . In what follows, express your answers in terms of m1 , m2 , ! , µ k and the acceleration of gravity g (or some subset of these parameters). d) Draw separate free body diagrams for the second (freely suspended) pulley, and the block on the inclined plane, showing and labeling carefully, all the forces that act on the objects. e) Find the magnitude of the acceleration of the block on the incline plane. Solution: We shall first assume that if there were no friction the block would slide up the incline plane. Therefore the static friction force points down the inclined plane. There is no acceleration so applying Newton’s Second Law for the block on the incline plane yields N ! m1 g cos" = 0 (1) T ! f s ! m1 g sin " = 0 (2) . Applying Newton’s Second Law to the system consisting of the object hanging from the suspended pulley plus the suspended pulley yields m2 g ! 2T = 0 (3) We can solve for the magnitude of the static friction. Solve Eq. (3) for the tension T , and substitute that result into Eq. (2), and solve for magnitude of the static friction: fs = 1 m g ! m1 g sin " 2 2 (4) Note that the maximum value of static friction is f s,max = µs m1 g cos! (5) We can find the condition on the masses such that the block just slips up the inclined plane m2 > 2m1 ( µs cos! + sin ! ) (6) Now suppose the block is sliding up the incline plane T ! f k ! m1 g sin " = m1ax1 (7) f k = µ k N = µ k m1 g cos! (8) . The kinetic friction is given by Thus T ! µ k m1 g cos" ! m1 g sin " = m1ax1 (9) The force equation on block 2 plus the attached pulley is m2 g ! 2T = m2 a y 2 (10) Constraint: The length, call it L , of the string is fixed. Let R be the pulley radius. From our coordinate system we have that L = (D1 ! x) + " R / 2 + y + " R + (D2 + y) (11) After two derivatives with respect to time gives 0 = !ax1 + 2a y 2 (12) ax1 = 2a y 2 (13) Thus Solve Eq. (9) for T and substitute the result into Eq. (10), m2 g ! 2( µ k m1 g cos" + m1 g sin " + m1ax1 ) = m2 a y 2 (14) Then use Eq. (13) yielding # m ! 2m1 ( µ k cos" + sin " ) & ax1 = 2 % 2 (g 4m1 + m2 $ ' (15) When m1 = 0 block 2 is in free fall with an acceleration g , which makes sense. This result is positive as long as m2 > 2m1 ( µ k cos! + sin ! ) . When m2 = 2m1 ( µ k cos! + sin ! ) the tension in the string just balances the friction force agreeing with our result in Eq. (6) provided we assume that µs = µ k . Because µs > µ k , when the block just slips, it will immediately slide because the kinetic friction force is smaller than the static friction force. If we want to find the acceleration when the block on the incline plane slides downward then we must reverse the direction of the kinetic friction. Thus Eq. (9) becomes T + µ k m1 g ! m1 g sin " = m1ax1 (16) Otherwise the equations are the same hence ay 2 = m2 ! 2m1 (! µ k cos" + sin " ) g 4m1 + m2 # m ! 2m1 (! µ k cos" + sin " ) & ax1 = 2 % 2 (g 4m1 + m2 $ ' (17) (18) When m2 = 2m1 (! µ k cos" + sin " ) the gravitational force just balances the tension in the string and the friction force. For m2 < 2m1 (! µ k cos" + sin " ) , ax1 < 0 and the block slides down the inclined plane. Problem 9: Winter Sports Calvin and Hobbes are riding on a sled. They are trying to jump the gap between two symmetrical ramps of snow separated by a distance W as shown above. Each ramp makes an angle ! with the horizontal. They launch off the first ramp with a speed vL . Calvin, Hobbs and the sled have a total mass m . a) What value of the initial launch speed vL will result in the sled landing exactly at the peak of the second ramp? Express your answer in terms of some (or all) of the parameters m , ! , W , and the acceleration of gravity g . Include in your answer a brief description of the strategy that you used and any diagrams or graphs that you have chosen for solving this problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Answer: The sled is undergoing projectile motion with an initial launch speed vL in the direction given by a positive ! with respect to the horizontal. We choose a coordinate system as shown in the figure below: Then the components of the acceleration are given by ax = 0 and a y = ! g . We are given the range W , so we can use the equations for the y-components of position to find the time of flight and then use the x-component of position to find the initially launch speed vL . The y-component of the position is given by 1 y (t ) = vL sin ! t " g t 2 2 (1) Since the sled returns to it’s initial starting point at the instant t = t f , Eq. (1) becomes y (t = t f ) = 0 = vL sin ! t f " 1 g tf 2 2 (2) We can solve Eq. (2) for the time of flight tf = 2vL sin ! g (3) The x-component of the position is given by x(t ) = vL cos ! t (4) At the end of its flight. Eq. (4) becomes x(t f ) = W = vL cos ! t f (5) Substitute Eq. (3) into Eq. (5) to find that 2vL 2 cos ! sin ! vL 2 sin 2! W= = g g (6) We can now solve Eq. (6) for the launch speed vL = gW . sin 2! (7) b) Hobbes was clutching a bag of tiger food when they left the first ramp. He got so excited while in the air that he let go of the bag at the top of the flight, lightening the total mass attached to the sled by 10%. Explain qualitatively how this will affect the results you found in a). Answer: When we apply Newton’s Second Law to the sled, noting that the gravitational force is proportional to the mass, the mass cancels out and the acceleration is independent of the mass. Therefore when Hobbes releases the bag, the acceleration does not change, hence the trajectory remains the same. (Note that if Hobbes threw the bag in any direction with some speed then the trajectory would change do to recoil; a process will study shortly.) c) Explain qualitatively what happens to the released bag of tiger food. Answer: Since the bag of food was released from rest, it’s trajectory is the same as the sled. d) There is friction between the sled and the snow which can be modeled by coefficient of kinetic friction µk . What is the minimum value of µk that will assure that the sled comes to rest somewhere on the second ramp? Assume an infinitely long ramp. Include in your answer a description of the plan that you will use to solve this part of the problem and any relevant diagrams or graphs that you find useful. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Answer: In order for the sled to eventually come to a stop (on an infinitely long hill) the acceleration must point uphill. We draw a free body force diagram with directions chosen as shown in the figure below. Since the friction force points uphill, we can apply Newton’s Second Law to find a condition such that the acceleration up hill is less than zero. The forces on the sled (and riders) are gravity ! mg = mg( î cos ! " ĵsin ! ) (8) and the contact force between the ground and the sled ! ! ! C = N + fk = N ˆj ! f k ˆi . (9) ! ! The components of the vectors in Newton’s Second Law, F = ma , are mg sin ! " f k = m ax N " mg cos ! = m a y . (10) (11) The condition that the sled slide down the ramp is expressed by a y = 0 and the model for kinetic friction is The condition a y = 0 in Eq. (11) gives f k = µk N . (12) N = mg cos ! , (13) f k = µk mg cos ! . (14) and so Eq. (12) becomes Substitute Eq. (14) into Eq. (10) to solve for the x -component of acceleration ax = g (sin ! " µk cos ! ) . (15) In order for the sled to come to rest, the x -component of acceleration must be less than zero ax = g (sin ! " µk cos ! ) < 0 . (16) Hence we arrive at the condition that tan ! < µk . (17) Problem 10 Bicycle and the Falling iPod A bicycle rider is traveling at a constant speed along a straight road and then gradually applies the brakes during a time interval 0 < t < t f until the bicycle comes to a stop. Assume that the magnitude of the braking force increases linearly in time according to F = bt , 0 < t < tf where b > 0 is a constant. The cyclist and bicycle have a total mass m . At the instant the person applies the brakes, a horizontal distance d from the rider, the wind blows and snaps an iPod off the branch of the tree with a horizontal speed v po in the direction shown in the figure below. The ipod was initially a height h above the ground. The cyclist catches the iPod at the instant the cyclist has come to a stop. You may assume that the cyclist catches it at a height s above the ground. a) At what time t1 did the cyclist catch the ipod? Express your answer in terms of the quantities, b , m , h , s , v po , d , and g as needed. b) What was the initial speed of the cyclist? You may leave your answer in terms of t1 from part a) and any other quantities as needed. Solution: There are two objects moving, the bicyclist and the iPod. We shall choose a coordinate system for the bicyclist and the iPod as shown in the figure below. We can use Newton’s Second Law to calculate the acceleration of the bicyclist. We can then integrate the acceleration to find the x-component of the velocity of the bicyclist. We then use the equations for free fall to determine the time it takes the iPod to fall a distance h ! s . The x-component of the velocity of the bicyclist is zero at this instant so we can solve for the initial x-component of the velocity of the bicyclist. a) At what time t1 did the cyclist catch the ipod? Express your answer in terms of the quantities, b , m , h , s , v po , d , and g as needed. ! ! From Newton’s Second Law, F = ma , are in the x-direction: ˆi : !bt = ma . x Thus the x-component of the acceleration is ax = ! (1) b t m (2) We can now integrate the x-component of the acceleration of the bicyclist to get the xcomponent of the velocity of the bicyclist t t b 2 ! b " vbx (t ) # vbx 0 = ( ax dt = ( $ # t % dt = # t . m 2 m & ' 0 0 (3) So the x-component of the velocity of the cyclist as a function of time is given by vbx (t) = vbx0 ! b 2 t 2m (4) The y-component of the position of the iPod is given by y (t ) = (h ! s ) ! g 2 t 2 (5) The bicyclist catches the iPod at the instant when y (t = t1 ) = 0 = (h ! s ) ! g 2 t1 2 (6) We can solve eq. (6) for the flight time: t1 = 2(h ! s ) g (7) b) What was the initial speed of the cyclist? You may leave your answer in terms of t1 from part a) and any other quantities as needed. The bicyclist comes to a full stop at the instant t1 so we can substitute Eq. (7) into Eq. (4) and solve for the initial x-component of the velocity of the bicyclist b 2 b( h ! s ) t1 = 2m mg vbx 0 = (8) Alternatively we can integrate Eq. (4) to find that the x-component of the position the bicyclist when it just came to a stop t1 t1 b 2" b 3 ! xb (t1 ) # xb 0 = ( vbx dt = ( $ vbx 0 # t % dt = vbx 0t1 # t1 . 2m ' 6m 0 0& (9) We chose x0 = 0 and so substituting x0 = 0 and Eq. (7) into Eq. (9) yields xb (t1 ) = vbx 0t1 ! b 3 t1 6m (10) During the time the iPod fell, it traveled a horizontal distance x p (t1 ) = v pot1 = v pot1 (11) xb (t1 ) = d + x p (t1 ) (12) So Substituting the results above yields vbx 0t1 ! b 3 t1 = d + v pot1 6m (13) vbx 0 = b 2 d t1 + + v po . 6m t1 (14) We can now solve for vbxo : Note: The parameters b , d , and v po are not independent. They are related as follows. We can integrate Eq. (4) using the value for vbx 0 that we found in Eq. (8) to find that the x-component of the position the bicyclist when it just came to a stop tf t f ! b( h # s ) b 2 " b 2" b( h # s ) b 3 ! xb (t f ) # xb 0 = ( vbx dt = ( $ vbx 0 # t % dt = ( $ # t % dt = tf # t f .(15) 2m ' mg 2m ' mg 6m 0 0& 0& t We chose x0 = 0 and so substituting x0 = 0 and Eq. (7) into Eq. (15)yields b ! 2(h # s ) " xb (t f ) = $ %% 3m $& g ' 3 (16) During the time the iPod fell, it travelled a horizontal distance x p (t f ) = v pot f = v po So 2(h ! s ) g (17) xb (t f ) = d + x p (t f ) (18) Substituting the results above yields 3 b ! 2(h # s ) " 2(h # s ) $$ %% = d + v po 3m & g g ' (19) We can now solve for the constant b in terms of the other constants and v po : ! 2(h # s ) " b = 3m $ d + v po % g & ' 3 ! 2(h # s ) " $ % . g & ' (20) Problem 11 Two Block Pull Consider two blocks that are resting one on top of the other. The lower block has mass M 2 and is resting on a nearly frictionless surface. The upper block has mass M 1 < M 2 . Suppose the coefficient of static friction between the blocks is µs . a) What is the maximum force with which the upper block can be pulled horizontally so that the two blocks move together without slipping? Draw as many free body force diagrams as necessary. Identify all action-reaction pairs of forces in this problem. b) What is the maximum force with which the lower block can be pulled horizontally so that the two blocks move together without slipping? Draw as many free body force diagrams as necessary. Identify all action-reaction pairs of forces in this problem. Solution: a) Suppose we take as our system both blocks. Then the free body diagram is shown below Applying Newton’s Second Law to the system yields î : F1 = ( M1 + M 2 )a ĵ : N G 2 ! ( M1 + M 2 ) g = 0 (21) Thus a = F1 / ( M1 + M 2 ) (22) However we have no knowledge of the internal forces between! the surfaces of block 1 and block 2, so we cannot determine whether or not the force F1 will cause them to slip relative to each other. ! In order to determine some condition F1 due to the fact that static friction cannot exceed some maximum value we must draw free-body force diagrams for the respective blocks as shown below. Note that the static friction force is the force that accelerates the lower block. The action-reaction pairs are: i. The gravitational force (magnitude M 1 g ) that the earth exerts on the upper block and the gravitational force (not shown) that the block exerts on the earth. ii. The gravitational force (magnitude M 2 g ) that the earth exerts on the lower block and the gravitational force (not shown) that the block exerts on the earth. iii. iv. ! The normal contact force N 21 that the lower block 2 exerts on the upper block 1 ! and the normal contact force N12 that the upper block 1 exerts on the lower block 2. ! The static friction force f21 , directed to the left in the figure, that the lower block ! 2 exerts on the upper block 1 and the static friction force f12 , directed to the right, that the upper block 1 exerts on the lower block 2. v. The upward contact force of magnitude N G 2 that the surface (ground, table, or other) exerts on the lower block and the downward contact force that the lower block exerts on the surface vi. Whatever or whomever is pushing the block will have a reaction force of magnitude F1 directed to the left. Applying Newton’s Second Law to the upper block yields î : F1 ! f 21 = M1 a1 ĵ : N 21 ! M1 g = 0 (23) where a1 is the acceleration of the upper block. Applying Newton’s Second Law to the lower block yields î : f12 = M 2 a2 (24) ĵ : N G ! N12 ! M 2 g = 0 where a2 is the acceleration of the lower block. The just-slipping conditions are that the accelerations are equal, a ! a1 = a2 , (25) and that the static friction has its maximum value, f 21 = ( f 21 )max = µs N 21 . We can now solve the above equations for the acceleration of the blocks. The first equation in (23) implies N 21 = M1 g (26) (27) and so Equation (26) becomes f 21 = µs M1 g . (28) Substituting Equations (28) and (25) into the first equation in (24), µs M 1 g = M 2 a and the acceleration of the blocks is a1 = a2 = a = µs M 1 g M = µs g 1 . M2 M2 (29) (30) The pushing force is then found by substituting Equation (30) into the first equation in (23), ! M " µMg (31) F1 = µs M 1 g + M 1 s 1 = µs M 1 g #1 + 1 $ . M2 % M2 & Equivalently, we could have substituted our result for the acceleration from Eq. (22) into Eq. (30) resulting in a = µs g M1 = F1 / ( M1 + M 2 ) M2 (32) which yields the same result as Eq. (31) for the magnitude of the maximum possible pushing force on the upper block before the blocks slip. Check this result in the limits that one block is much heavier than the other. If M 1 ! M 2 , the lower block won’t move at all, and F1 ! µs M 1 g , the maximum static friction force. If M 1 ! M 2 , the lower block acts more or less as a lubricant, and the maximum force becomes very large. b) The free-body force diagrams are shown below when we push the lower block, The static friction force is now the only force that makes the upper block accelerate. The action-reaction pairs are the same as in part a) (except for (vi), the reaction force of magnitude F2 on whatever or whomever is doing the pushing). We again apply Newton’s Second Law to each block. Applying Newton’s Second Law to the upper block yields î : f 21 = M1 a1 (33) ĵ : N 21 ! M1 g = 0. Applying Newton’s Second Law to the lower block yields î : F2 ! f12 = M 2 a2 ĵ : N G ! N12 ! M 2 g = 0. (34) The just-slipping conditions are identical to Equations (25) and (26). Substitution of these into the first equation in (33) yield a = µs g (35) After substituting Eq. (35) and Equation (25) in the first equation in (34), we find that the maximum pushing force is F2 = µs M 1 g + M 2 µs g = µs g ( M 1 + M 2 ) . (36) Note that if we draw a free-body diagram on the compound system, the free-body force diagram looks like Newton’s Second Law in the horizontal direction then becomes ( ) î : F2 = M1 + M 2 a (37) which is easily solved for the acceleration a= F2 . M1 + M 2 Then substituting Eq. (38) into Eq. (35) yields F2 a= = µs g M1 + M 2 (38) (39) which we can solve for the maximum pushing force as given by Eq. (36). We note by comparing Equation (30) to Equation (35) that the acceleration of the blocks is greater when you push the lower block than when you push the upper block (recall that the problem statement specified M 1 < M 2 ). Recall that the static friction force is the only force accelerating the block that you are not pushing so when you push the lower block ( f s )max ! M1a ! M1 F2 (push lower book) . (40) When you push the upper block ( f s )max ! M 2 a ! M 2 F1 (push upper book) . (41) The maximum value of static friction between the blocks has the same magnitude in both cases. Therefore the product M 2 F1 = M1 F2 (42) F1 M = 1 F2 M2 (43) or So M1 / M 2 < 1 , then F1 / F2 < 1 and you can push the lower block with a greater force F2 > F1 .