Question

PRINTER VERSION RCES BAC Chapter 3, Problem 3/013 X Incorrect Determine the tension P in the cable which will give the 54-lb block a steady acceleration of 6.9 ft/sec2 up the incline. 2 02 270 m. 54 lb Hi0.34 222 tudy 30.19886 Answer: P- the tolerance is +/-2% Click if you would like to show Work for this question: Open Show Work Activate Windows

Answer 1

Tension e in cable =? Mass of block 54 Lb 54 lb MK = 0.34 220 Acceleration 6.9 ft/sec) Accelera Hon due So, the block (ft /sec2) to gravity (9) = 32-2 iş on 220 Pncline. 54 - ab Let us draw the Free body Diagram: Fuß:0:- 8 AN ma Fan 54g sin (22) L 549 4b (220) 5489 forces acting self weight weight ( 5449) e

towards since body (i) friction force acting iş accelerating force (N) up- down [ f) iv) P (Tensile force in cable) 17) Reaction ~ Inertical Force (Fi= mal acting opposite of ine, down the to motion body Incline. forces (e) also resolve to need Tensile We So, :: Resolving in F-B.D them SP (08270) P sin (270) AN f Fizma (54X37-2) cs(22) 54432-2) sin(229) F-B:D

i) Since in body will be equations We equilibrium equilibrium can use of Sum of all horizontal force = 0 ZFH=0 P Cb (270)= f+ (54X32-2) sin(229) + Fi whore f = MAN where 4 k = 0.34 Fi mxa- (546 6:9) = 3426 pounds 3) Plo-891) = (0-34x0) + (54x32-2 X 0-3746) + 372.6 → 04 891 PE 0:34 N + 1023.96 > 0 vertical Sum of all forces =0 (Ef x = 0) P Sin (270) +N= (54x 32-2) Cos (220) 16120187 0°4539 PIN

Solving 0 & ② 0-891 P - 0034 N= 1023.96 16 12 - 187 7)%0.34 + N = 0.45 39 P 548143 7 0.1543P+ o fauna 0/340= 102 3.96 0.891P 100453 P = 1572.10358 = 1503-9 Pound-flat lov) lb-force- Pa 150304 Unit in Ib-fonly But Since should We have in lbs. pa 150309 answa but lb . = 46-7 32-2 1P = 46-7 lb XX