Question

Chapter 3, Practice Problem 3/020 With the blocks initially at rest, the force P is increased slowly from zero to 129 lb. Plot the accelerations of both masses as functions of P. Then answer the questions. 67 lb M = 0.43 i = 0.38 M = 0.38 = 0.33 90 lb P Questions: If P = 27 lb, AS ft/sec. as - ft/sec2 If P = 84 lb, DA= ft/sec2, as = ft/sec2 If P = 125 lb, aA = ft/sec2aB = ft/sec2

Answer 1

Ans: A 67eb Hy=0-43 Mg=0.387 Hy=0.38, HR -0.335 B Summation of forces AP qolb block A, m ΣΗ=0, 0 FF, AB MARA 20. [1=0 NAB-mag=0: MAI NAB Maximum Static friction is = 10.43) (67) FF, AB, =Hs, AB mag ,AB, max = 28.81 lb. maximum acceleration of block A without slip is FF, AB, maa = da,max, no slip ma 28.8) = 13.85. sito 52 67 lbm 132.2 ebm slug kinetic friction force between the block is FF, AB, slip Mc, AB Mag - (0.38) (67) = 25.46 Summation of forces block B, Elt=o (2) P- FF,AB -FEB-mäß = 0 Σv=0 =0, No-NaB-meg NB = NABt meg CS Scanned with CamScanner

maximum Static friction force is edotando FF,B,max = 45,8 HS B (MAX mog - 10:38) C67+90) 259.66 lb Kinetic fiction force is, (btw B. and ground) Fremax FF, B, slip HK,8 P (ma+mg) g =(0.337(67+902 = 51.81eb As is increased there will be no motion until P reaches robod 59.66 lb ~ bolb at which block B will begin to Slide along the lower Surface , considering equations O&O FF, AB EMARA O 2 = P- FF,B FEAB mis Q113 (3 maa A P- FF,8 - MBAB block B and the ground. There is slip between (MAI MB) g FF, B MIC, AB = 0.38 (67 +902 59.66 lb - Assuming there is no slip between blocks then aA = 98 (MA+MB) =P- FFB From 6 ay aB P- 59.66 aB From (4) Y MATMB P-59.66 AA AB P-60 4:85(slug 157132.33 CS Scanned with CamScanner

If P = 272b, QA = Op = Ofts- 84-60 If P = 84lb, a A = AB 4.95 FULS 4.85 125-60 If P= 125 eb AA - ab = 13.471s 4.85 CS Scanned with CamScanner
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