Antwoorden_Tentamen_Statica_Januari_2011

# S bc 15 kn trek s bf 25 kn trek scf 21 2132 kn

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Unformatted text preview: SCF cos 45° − 10 = 0 SCF = −21, 2132 kN Antw. S BC = 15 kN (trek) S BF = 25 kN (trek) SCF = −21, 2132 kN (druk) Pag. 2/4 Uitwerkingen tentamen Statica-1 (wb1114) 24.01.2011 Opgave 3 AB = 2 2 + 2,82 + 32 = 4,5651 m Richtingscosinussen van AB: 2 cos α = = 0, 4381 AB cos β = 2,8 = 0, 6134 AB cos γ = 3 = 0, 6572 AB ∑ M y = 0 : G ⋅1 − Tz ⋅ 2 = 0 Tz = 300 N Tz = T cos γ : T = 456,5085 N Antw. T = 456,5085 N ⎧Tx = T cos α = 200 N ⎪ ⎨Ty = T cos β = 280 N ⎪ ⎩Tz = 300 N C y = 0, 6Ty = 168 N ∑(M ) = 0 : C x ⋅ ( 2,8 − 0,8 ) + Tx ⋅ 0, 4 − Ty ⋅ 2 = 0 ∑(M ) C x = 240 N = 0 : G ⋅ (1, 4 − 0, 4 ) + Tz...
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## This document was uploaded on 02/26/2014 for the course MECH 2300 at Delft University of Technology.

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