Help with having an explination to understand the steps to this tension problem.
ID: 1600858 • Letter: H
Question
Help with having an explination to understand the steps to this tension problem. I solved this problem myself and attempted to explain the rational behind each step and showed how I calculated if we needed cos and sin , but I'm not sure if it is right or if it makes sense, so I would appreciate if you could look over my work, and if there are some wrong explanations please tell where and what the right explanation is. (The first picture is the question and my teacher's work, and then the pictures of my work which I need help with are after that.)
EDIT: Took the pictures again. If you still can't see them well right click on the picture and open image in new tab, and increasing your device's brightness may help too.
Thank you!
First Condition for Equilibrium Chandelier cord tension Calculate the tensions and F in the two cords that are connected to the 960 N vertical cord supporting the 200-kg chandelier shown. Ignore the mass of the Cords First Condition for Equilibrium Solution Chandelier cord tension Note: F FB W 0 Fax FB FAY mg 1960N FA sin (60°) 1960N 3 FA 2260N Fe E FACos (60 s) E 1130N First Condition for Equilibrium Solution Chandelier cord tension Note: FA FR W 0 60 mg 1960N AX AY FA sin (60°) 1960N 3 FA 2260N 1960 N FB FAcos(60 o) 1130N 200 kg Slide 18 of 22 Solving Statics Problems Ay Solution: The forces at the point where the three cords join must add to zero. The vertical component of FA is the weight; the horizontal components of FA and FB cancel. Therefore, F 2260 N and F 130 N.Explanation / Answer
resolving the component of FA, WE get
ALONG X :FAX = FB and along y: FAY = W
or, FA sin60 = mg = 200x9.8 = 1960
or, FA = 2263.21 N
so, FAX = FB
or, FA cos60 = FB
or, FB = 2263.21 x cos60 = 1131N
they did it right.. urs is right also
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