Physics help needed here!!!!!!!!
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hi, i really need you guys to help me to solve this, its driving me mad!
Question 1http://i241.photobucket.com/albums/ff164/schmaltx/lols.jpg
(a) Compute and draw the reaction force in the ceiling which is holding the pulley
(b) If all the wires used to hold thes weights are of the same type and having a cross sectional diameter of 6mm. What shall be the minimum strength of these wires if a safety factor of 2 is to be adopted to hold these weights? (Ans: 128.2N/mm^2)
just in case you need to see the ans for F1-F6 & W
F1 - 500N
F2 - 700.1N
F3 - 1710.1N
F4 - 1330.22N
F5 - 1330.22N
F6 - 535.61N
W - 1812.37N
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Question 2: A strut AB is added to stabilise the hanging poles against the turning effects of the applied forces. Calculate the compressive force in the strut AB (Ans: Fig6a: 23.2kN, Fig6b: 13.91kN)
http://i241.photobucket.com/albums/ff164/schmaltx/HAHAH.jpg
just in case you need the answer for the bending moments about pivot at P for both figures, here's the ans, 6(a) - 58kNm cw , 6(b) - 40kNm acw -
there's no way I can answer at the moment because my office blocks photobucket and most other photo sharing sites
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Resolving forces, sianz 1/2.
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okay yes! these are the pictures, thank you maurizio13!
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diameter = 6mm
area = 28.27mm^2
largest force per unit area = 1812.37/ 28.27
include safety factor 2 = 128.2
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moment = force x perpendicular distance from pivot, which is P in the diagram.
4kN x 5 m +2m x 5kN + 4m x 7kN = 5m x F
F = 1.16N ===> for the x componentfinal F =1.16/cos 60
=2.32 -
Originally posted by skythewood:
diameter = 6mm
area = 28.27mm^2
largest force per unit area = 1812.37/ 28.27
include safety factor 2 = 128.2
okay, thanks. but i still don't get how come is using the largest force per unit area which is using the force "W"?
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5kN x 8m +F x 4m = 20kN x 4m
F = 10, x componentFinal F = 10 / cos 70
= 2.92 -
Originally posted by schmaltz:
okay, thanks. but i still don't get how come is using the largest force per unit area which is using the force "W"?
i have no idea. at first glance, i thought it should be use 2500N. but that didn't get your answer, so i used the next heaviest one.
I stongly suggest you double check the answer source.
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Originally posted by skythewood:
5kN x 8m +F x 4m = 20kN x 4m
F = 10, x componentFinal F = 10 / cos 70
= 2.92
hahaha, omg. den how do you get this answer? and what is x component?
my answer for this part is 10.64kN i double checked, the other one you asked me to double check, indeed is wrong, but is also different from your answer, the new answer i've got is 12.0N/mm^2 -
I used simple moments, clockwise wise, anti clockwise force...
than get 40kNm clockwise.
i use vertical height of 4m, and force in the horizontal side to get.... think i made some mistake there...
or you can just use the length of the strut. 40kNm = F x 4.26m
F = 9.4012.0N/mm^2. wow. you will need a cross section area of 208mm^2 to hold 2500N. include the safety factor, and it will be 416mm^2. your diametre will than be 23mm.
guess i should retire... can't even solve moments le....
also check the 6a) bah, i did that one quite rashly too...
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anyone has different solutions to this?
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6a) Pivot is at P (to prevent turning of poles)
Let F = compressive force
Anti clockwise moments = F cos 60 * 5 = 2.5 F
Clockwise moments = 4 kN * 5 + 5 kN * 2 + 7 kN * 4 = 58 kNm
Hence,
2.5 F = 58 kNm
F = 23.2 kN
6b) Pivot is at P (to prevent turning of poles)
Let F = compressive force
Anti clockwise moments = 20 kN * 4
Clockwise moments = F cos 70 * 4 + 5 kN * 8
Hence,
20 kN * 4 = F cos 70 * 4 + 5 kN * 8
4F cos 70 = 40 kNm
F = 29.2 kN -
skythewood is correct in his analysis for the other question.
Recheck your answer source.
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okay sure, but anyway thanks for all the help! =D
really much appreciated~