Question

Mechanical Engineering Departement [1] If the bucket and its contents have a total weight of 45 N, determine the force in the supporting cables DA, DB, and DC. 2.5 mm 4.5 mm 3 1.5 mm D 1.5 mm

Answer 1

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(31-1.5g+3 kg (4.5,013) 4 be the cables. DA ID (1.5, 1.5, o) FDB posinon The solution is derived using Vector approach. Let FORFA & Foc $0,25,3) magnitude of forces in the (165,0,0) Let OA ,OB, oc & OD are the vectors of points A,B,C & D. 45N. Then 4.5° toj +3K, = 1.5i to og OC = 0% +2.50 +3 k., 1.5 itlat 105 g tok given by , DA= O 31 -1.59 +3K. oi -155 i tok. Similarly -105j = 7.50 +1j +3K OA = tok. Theis DA is -- OB OD DB DC Ос -OD DA is Now unit vechor along A DA U DA (20.25 given by load √ 3²+1.52+32 magninde oi-1.5g tok oi- j tok a 17:52 -1.5i + 15 +3k (-1-58+13 +36) $1.5² + 2+ 32 ^ DB u "DB IDBI Similarly DC 12.25 De ibct bg mulbplying The cable forces can be Converted to Vector form with respective unit vectors. FDA (38_1.55+36) 20.25 - FDB D C For FB (oi-gtok Foc (-1158 +1; +36) ViQias z direchas. Therefore it's Force act along vector representation will be -45k. ve 45N

Sum of these forces will be zero At equillibrum There fore Sum of a Components ie FDA x 3 + Foc X 1.5 (20.25 12.45 FDA 3 X X 112.95 Грс 20-25 1.5 1.555 FDA DC Sum of § Components = 0 in. FDA x1.5 20.25 FDB. + Foc J12.25 .105 FDA + 2 FDB Foc J12.25 120.25 Sum of k Components -45 X 3 + FOA Jaoias FOC X3 Ji2.as frons eanahon ③ Foc substitute for 3 FDA 45 0 + 1.555 FDA 5 12.25 X3 do: 25 45 1.9995 FDA FDA 22.5 N / = 34.99 NG F be = 1.555 X 22.5 From equation 34.99 = 2.5NK FDB -1.5 X 2205 120-25 From cavalon 112.25