Question

ENGR 115 Week 5 Practice Exercise 11 Name AC, and AD SO 3. Determine the maximum weight of the crate that can be suspended from cat that the tension developed in any one of the cables does not exceed 250 lb. с 3 ft 6 ft 40 3

Answer 1

let tensions developed in the cables ABACA AD be TABITA 8 To respectively. To get the directions of the tension in the cable, the unit vectort along AB, ACA AD Should be known. co-ordinates of A = (0.6.0) co-ordinato of B = (-3,0,2) co-ordinata of co-ordinates of D = (0,9,47 AB = B. Ā: -1-6h to R TABA= -80 bytoch 2 IABI= 7 unit vector along AB AB= AB -3765+ IAB! -0.4299 -0.8575 ÃC = TĀ = 21-6 7+ 3h IACI = r2²+(-63² + ² = 7 unit rector along +0.286 AB AC 27434 TACI 0.286 î.0.857 h +0.429 Å Ас -

ÃO = o Ā= oî+ + 4 (AOI= Jo²+32+ q2 = along AD is AD unit rector .. AD = Á 01435 tu 1.6j+0.8k TADI 5 AD = oito 6ŷ to.si Now, Tension in Neight Weacts in Ê direction. AB = TAG CÁB) TAB (-0-4299 -0.857 +0.2862) cable

Tension in cable AC = TACIAC) TAC (0.2867 -0.8579 +0.42918) Tension in cable AD - TAO CAD) Tanlond to.ph) Neight -W on +(-0.857 TAB -0.8577 AC Ас Summation of all forces body is Ethat = (-0.429 TAB +0.286T Ditt to GTAD) t(0.286 TAB +0.429TACTO 8TA STADTW)

Since the body is in equilibrium thet = 0 0.429 TABt 0.286 TAC=0 0.857TAB - 0.857T + 0.6T AC Арго -W-o -0.286 TAB+ 0.429 TACTO STAD multiplying equation @ with (08) we get, 1.14 TAB + 1.16TA 0.8TAD C AC add which gives, 1.426 TAB + 1.569 TAC-W=0 Now, By multiplying eq ① with (a) wegel, - 1.426 TAB +0.951TAC=0 Adding we get og 8 eq ☺ 252 TAC-W=0 2.52 TAC From (G w TAC 2.52 0.397W in eq -2 TAC Substitute TAC TAB 0.265 W which gives Substolute TAB. TAC in TAO = 0.943N

TAB = 0.265 W TAC = 0.397 w TAO 0.943 W TAD is From these it is clear than highest cable It mentioned a maximum tension of donelop can 250 lb. TADE 250 lb 950 -) N = 265. 1 lbs. 0.943 965.lbs. weight of crate