3.5-Mg engine is suspended from a 500kg beam and hoisted by a crane which gives an acceleration of 5 m/s^2 when it has a velocity of 4 m/s.?
velocity will not be relevant 3.5-Mg = 3500 kg Assume bar AB is the one reported to have a mass of 500 kg. If we assume symmetry so that AD and BE both carry equal loads F = ma F(AD) = (3500/2)(9.8 + 5) F(AD) = 25900 N F(AD) = 26 kN If we neglect the mass of the chains, F(AC)sin60 = ((3500 + 500)/2)(9.8 + 5) F(AC)sin60 = 29600 F(AC) = 34179 N F(AC) = 34 kN As a form of exchange, please remember to vote a Best Answer from among your results. This is not the same as like or thumbs up which are also nice and do boost points, but only after asker awards initial points with their vote.
