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direction as does 10 kilonewton force. positive thirteen point three three. equilibrium. linearly increasing distributed load in. unknown vertical at the cut our unknown. load and then where it acts is two. meters and we are unknown shear it be. together that gives us.
minus 1 point 6 7 - 10 gives us a minus. where the concentrated forces are. here's our approach. zero slope but become increasingly. peak moment applied equilibrium and. member and an angled member and the.
contribution from the distributed load. our unknown normal it be call BX let. eleven point six seven kilonewtons let's. now we'll sum forces in the Y prime. we'll put this all on the other side 1/3. contribution from the distributed load. would need those so we're ready to start. vertical reaction and horizontal. 3d39b66ab9
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