Question

mz 12) Al 1 100kg car is pulling a 2350 kg trailer over a bridge. The bridge is uniform in density and broge has mass 5000 kg/ The only two supports on the bridge are the two pillars shown. The bridge is at static equilibrium. Call the two forces F, and Fe for the left and right pillars respectively. Respect standard sign conventions (as mentioned at the beginning of the exam a) Using a rotational equation with the axis of rotation at the right-hand pillar, find the force exerted on the bridge by the left-hand pillar 0.5m 1.5m na FL 15m 7 Fv - 100 215) 2350 )(30m) 2 15m 247500 + 2115000 Fr= 236 2500 N

b. find the force exerted on the bridge by the right habd pillar.

please explain concepts and procedure step by step. thank you

Answer 1

solution. Taking pivot at the right Crane & Can hand pillar Einet - 0 → Anticlee wise torgue - Clockwise fosique 23509 11009 5600g -15m -5m 47.5m 15 Antóclockwise torque = clockwise 27500 Og *7.5 +2350g xo5 = 1100gxi's g88 m/s² + F x 15 537500 +1175 - 1650 +15 F/ >>38675 - 1650 = 155/ F = 3702559 – 2468-34 N *9.8 = 24189-73N & Taking pivost at the left hand pillars Elnet-o Anticlockwise forque = Clockwise torque 5000g 2350g 1000g > FRx 15 = 50oog x 7.5+2350g x 14.5 +11008x1605 =15FR =(37500 + 34075+ 18150 g FR - 89 725 X88 = 58620 34 N Na -5m- € 75m 7.5m 14.5m 16.5m FR = 58620.34 N