p0310

# p0310 - 3.10. CHAPTER 3, PROBLEM 10 207 3.10 Chapter 3,...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.10. CHAPTER 3, PROBLEM 10 207 3.10 Chapter 3, Problem 10 Problem: Block B rests on top of Block A. The mass of Block A is mA = m and the mass of Block B is mB = 2 m. The coefficient of kinetic friction between sliding surfaces is µk . Beginning at time 3 t = 0, a constant force, F , is applied to Block A. (a) Determine the accelerations, aA and aB , for Blocks A and B, respectively. Express your answers in terms of F , m, µk , and gravitational acceleration, g . (b) If µk = 1 and F = 4 mg , how long will it take for the left sides of Blocks A and B to be 4 3 coincident if they are initially a distance L apart? .. . ............ ....................... . . .................. . . ....................... . ............ ........................................... .............................. ............. ............................................ . . . . . . ............................... . . ............................... . .. . . . . . . ............... .............................. . . .............................. . ............................... . ...... . ...................... . ............................... .................................. ................................................................................................................. ....... . . . . . . . . . ... .. .. .. .. .. ........... . ........... . . .... .. .. .. .. .. .. ......................................................... .. . ............................................................................................................. ........................................................................................................................ .................................................................................................... . . ..................................................................................... . .......................................................................................................................... .. . . . . . .......................................................................................................................... . . . . ......................................................................................................................... . . ........................................................................................................................ ................................................................................................................................ ... ........................................................................................................................... ... ........................................................................................................................ . . .......................................................................................................................... . ..................... . .................... . .. . . . .......................................................................................................................... . . ......................................................................................................................... . .......................................................................................................................... ......................................................................................................................... . . ...................................................................................................................................................................................... .............................................................. . ........................ . ......................................................................................................................... . ............ . ......................................................................................................................... . . ............ . . . .................................................................................................................................................................................................................................................. .. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ................................................................................................................................................................................................................................................... .. ............ ............................................................. ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... . . L B F A Solution: To begin, we must first draw diagrams indicating the forces acting on each block. As shown in the figure below, Block B is acted on by its weight, 2 mg , a reaction force from Block A, NB , and the 3 friction force, µk NB , which causes it to move. For Block A, there are three vertical forces, viz., its weight mg , the reaction force needed to support Block B, NB , and the reaction force from the surface, NA . In the horizontal direction, the applied force, F , pushes Block A to the right. Additionally, the friction force from the bottom of Block A is µk NA . Finally, the friction force from Block B, µk NB , also resists Block A’s motion. Note that the latter is an example of Newton’s Third Law, i.e., for every action of one object on another, there is an equal and opposite reaction. 2 mg 3 B mg µk NB µk NA .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ................................................................................................. ................................................................................................. ................................................................................................. ................................................................................................. ................................................................................................. ........................ µk NB NB NB A F NA (a) S nce he mo on of bo h b ocks s en re y hor zon a necessar y he ver ca forces on each b ock mus van sh So we have 2 NB − mg = 0 and NA − mg − NB = 0 3 Thus he norma forces are NA = 5 mg 3 and NB = 2 mg 3 We can choose he or g n of our coord na e sys em o be a any po n we w sh For he presen purpose e x = 0 a he ef s de of B ock A when t = 0 S nce pos ve x s o he r gh he d fferen a equa ons for 208 CHAPTER 3. FORCE AND ACCELERATION the motion of Blocks A and B follow from Newton’s Second Law, viz. m d2 xA 7 = F − µk NA − µk NB = F − µk mg 2 dt 3 2 2 d2 xB m 2 = µk NB = µk mg 3 dt 3 where xA and xB denote the positions of Blocks A and B, respectively. Thus, since aA = d2 xA /dt2 and aB = d2 xB /dt2 , the accelerations of the blocks are aA = (b) For µk = 1 4 F 7 − µk g m3 and aB = µk g and F = 4 mg , the accelerations are 3 aA = 4 71 3 g− g= g 3 34 4 and aB = 1 g 4 Substituting into the differential equations for motion for Blocks A and B, we have m 3 d2 xA = mg ; 2 dt 4 ˙ xA (0) = 0, xA (0) = 0 ˙ xB (0) = L, xB (0) = 0 21 2 d2 xB m 2 = m g; 3 dt 34 Solving for xA and xB tells us that xA (t) = 32 gt 8 and 1 xB (t) = L + gt2 8 The left sides of the blocks are coincident when xA (t) = xB (t), which occurs when 32 1 gt = L + gt2 8 8 Solving for t, we have t=2 L g =⇒ 12 gt = L 4 ...
View Full Document

## This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

Ask a homework question - tutors are online