I will type in the questions as appeared in the book so thatyou can solve it eas

Question

I will type in the questions as appeared in the book so thatyou can solve it easily. Please help me, because the concept isvery difficult compared to solving simple torque questions. Thankyou for your assistance.
A uniform 5.00 m long horizontal beam that weighs 315 N isattached to a wall by a pin connection that allows the beam torotate. Its far end is supported by a cable that makes an angle of53 degrees with the horizontal, and a 545 N person is standing 1.50m from the pin. Find the force in the cable Ft and the forceexerted on the beam by the wall R if the beam is inequilibrium. #1. (Read above problem) Rework the example problem above with axis of rotation passingthrough the center of mass of the beam. Verify that the answers donot change even though the axis is different.
#2 (Different than given problem) A uniform bridge 20.0 m long and weighing 4.00 x 10^5 N issupported by two pillars located 3.00 m from each end. If a 1.96 x10^4 N car is parked 8.00 m from one end of the bridge, how muchforce does each pillar exert?

A uniform 5.00 m long horizontal beam that weighs 315 N isattached to a wall by a pin connection that allows the beam torotate. Its far end is supported by a cable that makes an angle of53 degrees with the horizontal, and a 545 N person is standing 1.50m from the pin. Find the force in the cable Ft and the forceexerted on the beam by the wall R if the beam is inequilibrium. #1. (Read above problem) Rework the example problem above with axis of rotation passingthrough the center of mass of the beam. Verify that the answers donot change even though the axis is different.
#2 (Different than given problem) A uniform bridge 20.0 m long and weighing 4.00 x 10^5 N issupported by two pillars located 3.00 m from each end. If a 1.96 x10^4 N car is parked 8.00 m from one end of the bridge, how muchforce does each pillar exert?

Explanation / Answer

Please submit only one problem at a time (site rules)! Balancing torque about the point of contact with thewall 2.5 * T sin 53 = 2.5 * 315 + 1.5 * 545 T = 804 N Now balance vertical forces where F is the upward force ofwall on pivot point F + T sin 53 = 315 + 545 F = 218 N Now take torque about center of mass 218 * 2.5 = 545 * 1 545 = 545      so the result is thesame

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