Question

A block with a weight of 99.8 lbf is suspended from cord AC and two springs AB and AD. Determine the force in cord AC and the spring stiffness of spring AB if the system is in equilibrium in the configuration shown in the diagram below. Both springs have an unstretched length of 22.5 in., and the spring stiffness of spring AD is KAD = 3 lbf/in. As part of your answer, draw a free-body diagram of the system. Enter your answer for the force in cord AC with units in the box below on Blackboard, and box both of your answers on your paper. 29 in. 23 in. C 8 in. B D 16.5 in. 16 in. А W 7 in.

Answer 1

Solution 23in - -zgin c Sim ak 16in 16.5in given, We 99.8 lbf. unsketched length for bath spring (o)= = 22.5 in. Kag= 3 lbflin. FAC = ? KAB= ? K w 0 kring now, - tanco, = 16 ♡ = tant (I 16 30 23+7 28:07 6) also, sino, 16 16. TAD Lilao 34in Si228.07 tan02=8+16 .:0,- tamil ( 24 )= 73.74 7 = tans 03= 16.5 ...Oz fewesi ) = 36.869 also, Sinoza 16.5 TAB 29-7 AB= 16.5 • 27.50 .in in 36.86 now, extension in spring AB = x, = il AB-lo = 5 in extension in spring AD= x2 = lao-doz 11.5 in fibid of the systems and resolving all the forces along x-asis and Y-azus

Arsi 03 2 Fs.woso, al equilibrium qY FACSing + Fs, Sing, tfs, sina FAC condition Foll Efato ->X Fs, coso, + FACCUSO + FACCOSO, FS, COBO, w Fs, Co8O3 =0 Yct) KADA * co828.67 + (t) Vy' FACX (873-74 - KA BYX, C0836.86=0 sign cosmention → 10.147 KAD + 0.28 FAC = KABX 3-893 30.441 + 0.28 FAC = 3.893 KAB - (0) also, Efyzo A csino, & Fs, sino + Fs, sino -> 0.96fac + KADXX, sio 28.07 + kapx se, sin 36-86– 99.820 5) 0.96 FAC + 16.233 + KAB X 3 -99.870 TYLI - Wo 2) رنز) 0.96 FAC + 3 KAB = 83.567 now, lux0.96 (n,x0.28 gives 29.22 + 0.268FAC -(0.268fAC + 0.84 KAB) = 3.737 KAB 23.9 KAB=11.497 lbflin from equation Well Fac; 3.893x11.497 - 30.991 = 51.13 lbf. 0:28 KAB = 11.497 lbf /ios 51-13 lbf Aus- FAC