# Tension forces

1. Sep 17, 2006

### rwofford

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 Files:

• ###### steelcable.gif
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2. Sep 17, 2006

### rwofford

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?

3. Sep 17, 2006

### andrevdh

To convert from say 15 km to meters all you need to do is replace the kilo by $10^3$. So it becomes $15 \times 10^3\ m$ , as easy as that!

4. Sep 17, 2006

### rwofford

i still dont understand how to get the answer..i tried the way above but the answer is wrong...what am i missing?

5. Sep 17, 2006

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.

6. Sep 17, 2006

### rwofford

so by solving this problem should my final answer...the tension of the cable be 6087.9 N?

7. Sep 17, 2006

### rwofford

8. Sep 17, 2006

9. Sep 17, 2006

### rwofford

it should be attached...

#### Attached Files:

• ###### steelcable.gif
File size:
3.8 KB
Views:
86
10. Sep 17, 2006

These attachments usually don't work, as they don't right now.

11. Sep 17, 2006

### rwofford

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...

12. Sep 17, 2006

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.

13. Aug 15, 2008

### ritwik06

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