tension forces


by rwofford
Tags: forces, tension
rwofford
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#1
Sep17-06, 07:59 AM
P: 22
The steel I-beam in the drawing has a weight of 8.90 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? (picture attached)

for this problem do i need to first convert kN to N?...and im just not sure where to start...
Attached Thumbnails
steelcable.gif  
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rwofford
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#2
Sep17-06, 08:30 AM
P: 22
so the kN would equal 8900N and so to find the tension of the cables I would make a free body diagram. So there are two tension forces adn i label them t1 and t2.

F=2T cos 70
T=mg so T=8900(9.8)
F=2T cos 70=2mg cos70= 2(m)(9.8)cos70

but I dont know how to convert the kN...should i convert this to kilograms...how?
andrevdh
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#3
Sep17-06, 08:54 AM
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To convert from say 15 km to meters all you need to do is replace the kilo by [itex]10^3[/itex]. So it becomes [itex]15 \times 10^3\ m[/itex] , as easy as that!

rwofford
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#4
Sep17-06, 02:29 PM
P: 22

tension forces


i still dont understand how to get the answer..i tried the way above but the answer is wrong...what am i missing?
radou
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#5
Sep17-06, 02:46 PM
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Basically, [N] is converted to [kg] by dividing with g = 9.81 [m/s^-2], since m = F / a, and a = g = 9.81. F is your weight in [kN]. Convert to [N] by multiplying with 10^3.
rwofford
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#6
Sep17-06, 02:54 PM
P: 22
so by solving this problem should my final answer...the tension of the cable be 6087.9 N?
rwofford
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#7
Sep17-06, 03:59 PM
P: 22
someone please help...6087.9 isn't the right answer and ive tried this problem 1000 times!
radou
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#8
Sep17-06, 04:12 PM
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Putting the image on a link would be helpful.
rwofford
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#9
Sep17-06, 04:38 PM
P: 22
Quote Quote by radou
Putting the image on a link would be helpful.
it should be attached...
Attached Thumbnails
steelcable.gif  
radou
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#10
Sep17-06, 05:17 PM
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These attachments usually don't work, as they don't right now.
rwofford
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#11
Sep17-06, 05:24 PM
P: 22
well it is a picture of a crane that has a string coming down that splits into two cables one connecting on one side of the beam and the other connecting to the other side of the beam. A 70 degree angle is formed between one side of the beam and the string and likewise on the other isde of the beam...
radou
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#12
Sep17-06, 05:37 PM
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Okay, constant velocity implies equilibrium, firstly. First we write the equation for the, let's say, x-direction: T1cos70 - T2 cos70 = 0 => T1 = T2 = T. So, we prooved what was obvious - the forces in the strings are equal. Now we write the equation for the y-direction: 2Tsin70 - W = 0, where W is the weight of the beam, expressed in [N] od [kN], it doesn't matter at all, so just plug in your initial number from the text. From this equation we get T = W / (2*sin70). I hope this will help you.
ritwik06
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#13
Aug15-08, 08:39 AM
P: 586
Quote Quote by rwofford View Post
well it is a picture of a crane that has a string coming down that splits into two cables one connecting on one side of the beam and the other connecting to the other side of the beam. A 70 degree angle is formed between one side of the beam and the string and likewise on the other isde of the beam...
Well, if I am not wrong the answer if 4735.59 N Is it not?

The weight is alreay given in nwton. Why are you worried about about the mass (m). The earth pulls th iron beam with a force of 8900 N and the tension in the two cables pull it up. As there is no acceleration(uniform velocity), the sin components of each tension will balance the "weight". And as is seen from the FBD, the cos components will cancel out.

Ritwik


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