Question

Dynamics, please answer all parts.Thank you

Given: The mass of block C (mc) is 30 Lb. The mass of block A (mA) is 50 Lb, and the mass of block B (ms) is 75 Lb. You may neglect the masses of the pulleys, and you may assume that the blocks are initially at rest. Find: a) The equation(s) for position, velocity, acceleration, and displacement b) The displacement of the block at Bif A is pulled down a known distance of 3 ft. c) If the friction coefficient (px), between block C and its support surface, is 0.10, find the acceleration of each block (i.e., a4, as, and ac). d) The tension (F) in the cord which corresponds to the frictional condition in part (c). e) The position of block A as a function of time. f) The velocity of block A, B, and C as a function of time. g) Plot on the same graph the velocity of blocks A and B as a function of time. h) Show graphically how the block velocity changes as the friction coefficient varies from 0 to 0.5. B Solution

Answer 1

ay Tensions are shown in diagram, 44 k xc fe T assuming Massless pulleys. Free Body diagrany : 427 A 47 ac ac XB- TA TA T SA А an B Last С T T mag MB 9 2T NOTE :- A time. The direction of acceleration have been assumed and negative value of any one of them would mean that the actual direction of motion is opposite to assumption Now, the constraint in the motion is that the length of the streng is fixed. (1) ► 284 +48LB + ble =0 Equation of posse lors) differentiating Wirt ci) & QUA +40B + &c=0 (Equation of velocities). differentiating Wort time again. cili) 204 + 498 + ac=0 (Equation of acceluation) (V) ma9-27 MADA 32 #4182 (V) = MBS - 4T = тв ав ma= 5016 given ci) T= mcac MB = 75 lb) mc = 30eb. (Enough equations for saring all unknowns): đô vuông AA& ab in cill) from (V) &V), = os valus. (6) 29- 4T МА + 49 - 167 tac mo =0 Now, substitute lac tum cris in bor Egn. me 69 y me 16/mc málae - Io como en la act ac=0 69 - 12 ac - stac fac

69 = saya ac = ac = 6 X 32 X 5 39 ft/5² » 4 = 24.62 ft/s? - © .: Tomex ac = 738.46 26.ft/52 2 .. from (iv) & ② aA = g- 2T = 32- 29.5 ma Ә(ад = 2.5 ft/5? 3 - flom (v) & 4T aB g- - 32-39.5 mo AB 7.5 ft/529 » [aa t] [as A] tac (clirections). Therefore, ♡ A is pulled down by a eh pulled up is С prilled towards left.