A uniform 5.58 m long horizontal beam that weighs 278 N is attached to a wall by

Question

A uniform 5.58 m long horizontal beam that weighs 278 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 54? with the horizontal, and a 488 N person is standing 1.3 m from the pin. Part 1: Find the force FT in the cable by assuming that the origin of our coordinate system is at the rod

Explanation / Answer

Similar question with different values just change the values Question: A uniform horizontal beam 5.78m long that weighs 492 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 55? with the horizontal, and a 402 N person is standing 1.81 m from the pin. ANSWER Moments first Mo-Person = 402 N * 1.81 m = 727.62 N*m CW Mo-CG = 492 N * 5.78/2 = 492 N* 2.89 m = 1421.88 CW Total CW Moments: 727.62 + 1421.88 = 2149.5 N*m CW We need 2149.5 N*m CCW to balance that ===> Vertical component of cable * 5.78 m = 2149.5 N*m FVX = 2149.5 / 5.78 = 371.88 N UP Now the trig. The cable is the hypotenuse and Fvc is the opposite side sin 55 = 371.88 / T 0.819 = 371.88/T T = 371.88 / 0.819 = 453.98 N ==> Tension in cable The horizontal component of cable is the adjacent side cos 55 = Fhc / 453.98 Fhc = 260.39 N ===> Toward wall (Left) Sum vertical forces (down -) Person + CG + Fvc + wall Reaction = 0 -402 +(-492) + (+371.88) + wall reaction v = 0 -402 - 492 + 371.88 + WRv = 0 -894 + 371.88 + WRv = 0 WRv = + 522.12 N (UP) Sum Horizontal Forces (Right +) - Fhc + WRh = 0 -260.39 + WRh = 0 WRh = + 260.39 (Right) Tension in cable:453.98 N Forces in Pin: Fx = +260.39 N (Right) Fy = + 522.12 N (Up)

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