Is the Calculated Tension of 223 N Correct for the Clothesline Problem?

[Windwaker2004]

A clothesline is tied between two poles, 8 m apart. The line is quite taut and has negligible sag. When a wet shirt with a mass of 0.8kg is hung at the middle of the line, the midpoint is pulled down 8 cm. What is the tension in the clothesline at this time?

This question is relatively easy and I know you use trig mainly for the triangle and solve for the tension. I worked it out and got 223 N for the tension. Can anyone confirm this please?

 

[christinono]

I tried solving this problem with trig, but I got something else. Could you show me how you arrived to your answer?

 

[Windwaker2004]

I made a right triangle with a side of 0.08 m and a hypotenuse of 4 m (half of the clothesline) and solved for the angle to the horizontal. Once I got the angle it's just simple trig because the downward force is 7.8 N (0.8kg x 9.8 m/s^2). BTW I got the angle to be 1 degree.

 

[christinono]

wait 1 sec, I'm trying to figure it out...

 

[christinono]

Sorry I took so long, but my computer jammed (I hate this machine...).
Anyways, I got 392N for the tension force. I drew a vector diagram (Fg pulling down, one side of the string pulling down and to the right, and the other side of the strind pulling down and to the left.
Then I found the angle, which is equal to about 1 degree, and made a triangle with a hypotenuse of x (the tension force), an angle of 1 degree and an adjacent side of half the Fg (7.84N /2). I got 392N for x.

 

[Windwaker2004]

Ok i'll try to redo it and see what i get

 

[Windwaker2004]

Ok I got 392 N too but I had to multiply my answer of 196 N by 2. I got 196 N for the tension in one side of the rope. So does this mean 392 N is the total tension?

 

[christinono]

I'm pretty sure you have to multiply it by two, unless the answer specifically asks to find the tension one one side of the rope.

 

[Windwaker2004]

Oh and I just realized I didn't even use the 1 degree angle to solve for tension. I just used similar triangles.

 

[christinono]

Actually, according to my calculations, 392N is the tension in each side of the rope. The total tension would thus be 784N.

 

[Doc Al]

The clothesline is the hypotenuse of a right triangle, the vertical side is 0.08 m, the horizontal side is 4 m. The tension in the line makes an angle of $\theta = \arctan(4/0.08)$ with the vertical. The vertical components of the tension from both sides of the line must balance the weight of the shirt: $2T cos\theta = mg$. Solve for T.

 

[Windwaker2004]

Yea I just used your equation Doc and I got 196 N. So does this mean its 196 N of tension in each side of the rope or total?

 

[christinono]

Actually, if I'm not mistaken, if you want to use the formula: 2Tcos(theta)=mg, theta will equal 90-1.1457...degrees, which is approx. 88 degrees.

 

[christinono]

Sorry, made a mistake. Doc's method is right, just like my original method.

196 is the tension in each side of the cord.

 

[Windwaker2004]

yea that 88.85 degrees would be to the vertical so it still works.

 

[Windwaker2004]

Alright thanks again!

 

[Doc Al]

Yea I just used your equation Doc and I got 196 N. So does this mean its 196 N of tension in each side of the rope or total?

The tension is uniform throughout the rope. So there is no meaning to "total" tension. You are confusing the tension (a property of the rope) with the force that the rope exerts on the shirt: which gets a contribution from both sides of the rope that help support the shirt.

 

[Windwaker2004]

Ok thanks for straightening me out. But one last thing. You know how there is 8 m between the poles? Well this must mean that the rope is also 8 m because it is quite taut. But when it moves downwards the horizontal is still 4 m but why would the length of the clothesline (hypotenuse) change?

 

[christinono]

We never assumed the length of the cord changed.

 

[Windwaker2004]

True but if u calculate the hypotenuse using the sides 4 m and 0.08 m there is a little difference.

 

[Windwaker2004]

I guess it is negligible because it is such a small difference.

 

[Doc Al]

Ok thanks for straightening me out. But one last thing. You know how there is 8 m between the poles? Well this must mean that the rope is also 8 m because it is quite taut. But when it moves downwards the horizontal is still 4 m but why would the length of the clothesline (hypotenuse) change?

I make the reasonable assumption that the rope stretches before the poles move closer. But something's got to give.

For small sag it won't make much difference.

 

[christinono]

I don't quite understand. Are you talking about when we figured out the angle, because when we were doing so, the hypotenuse was 4 and the opposite side was 0.08. This way, we did not assume the cord stretched.

 

[Windwaker2004]

Yea I agree. It makes sense now lol. OK thanks to the both of you!

 

[christinono]

no problem

 

[Windwaker2004]

Yes but if you don't assume the cord stretches, this must mean the poles give in a little or the cord just wouldn't budge. But like Doc said, I guess its more reasonable that the cord stretches before the poles move in.

 

[christinono]

yeah, I see what you mean now...

 

[Doc Al]

Yes but if you don't assume the cord stretches, this must mean the poles give in a little or the cord just wouldn't budge. But like Doc said, I guess its more reasonable that the cord stretches before the poles move in.

Just for fun, figure out the angle both ways and compare. The way I did it, $cos\theta = 0.019996$; the other way gives $cos\theta = 0.02$. No measureable difference!