0.560 m CoM 0.300 m 13. The diagram above shows a NON-uniform beam of length 5.7
ID: 1997693 • Letter: 0
Question
0.560 m CoM 0.300 m 13. The diagram above shows a NON-uniform beam of length 5.70 m. The beam is held horizontal by two supports Support "A" is beneath the beam 0.860 m from the left end. Support “B" is above the beam 0.300 m from the left end. Support "A" applies an upward force of 482 N to the beam. Support "B" applies a downward force of 306 N to the beam. What is the distance between support "A" and the beam's center of mass (CoM in the diagram)? a. 2.85 m b. 1.99 m C. 2.29 m j. l'm getting an answer that is not one of the choices listed above and I'm sure that my work is correct. Please d. 1.21 m e. 0.841 m f. 0.918 m h. 0.974 m i. Zero grade the work that is neatly presented in my green book for this problem. I don't know how to work this problem. Please take this problem as being only 1 point rather than 2 points k. 14. The diagram above shows a drawbridge. The drawbridge is uniform, has a length of 3.25 m, a mass of 862 kg, and has a frictionless pivot at one end. The drawbridge is held in equilibrium by a rope attached 2.55 m from the pivot point. At the moment pictured the rope is horizontal and the drawbridge makes an angle of 24.0° relative to horizontal. What is the tension in the rope? a. 8450 N b. 9430 N C. 3440N j. l'm getting an answer that is not one of the choices listed above and I'm sure that my work is correct. Please d. 20800N e. 12100 N f. 7720 N g. 9250 N h. 18300 N i. Zero grade the work that is neatly presented in my green book for this problem I don't know how to work this problem. Please take this problem as being only 1 point rather than 2 points k.Explanation / Answer
13. balancing torque about the left end :
- (0.30 x 306) + ((0.300 + 0.560) x 482) - ((0.300 + 0.560 + d) ( W)) = 0
- 91.8 + 414.52 - (0.86 + d) W = 0
balancing forces in vertical,
482 - W - 306 = 0
W = 176 N
putting in previous equation,
- 91.8 + 414.52 - (0.86 + d)(176) = 0
d = 1 m 0.974 m
ANa(h)
14. balancing torque,
(3.25 / 2) (862 x 9.8) (cos24) - (2.55)(T) (sin24) = 0
T = 12100 N
Ans(E)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.