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Can someone confirm my answer please?by Windwaker2004
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#1
Sep2304, 04:41 PM

P: 34

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? 


#2
Sep2304, 05:59 PM

P: 213

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



#3
Sep2304, 06:17 PM

P: 34

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.



#4
Sep2304, 06:24 PM

P: 213

Can someone confirm my answer please?
wait 1 sec, I'm trying to figure it out...



#5
Sep2304, 06:40 PM

P: 213

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. 


#6
Sep2304, 06:45 PM

P: 34

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



#7
Sep2304, 06:54 PM

P: 34

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?



#8
Sep2304, 06:56 PM

P: 213

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.



#9
Sep2304, 06:57 PM

P: 34

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



#10
Sep2304, 07:00 PM

P: 213

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



#11
Sep2304, 07:00 PM

Mentor
P: 41,475

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 [itex]\theta = \arctan(4/0.08)[/itex] with the vertical. The vertical components of the tension from both sides of the line must balance the weight of the shirt: [itex]2T cos\theta = mg[/itex]. Solve for T.



#12
Sep2304, 07:04 PM

P: 34

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?



#13
Sep2304, 07:06 PM

P: 213

Actually, if I'm not mistaken, if you want to use the formula: 2Tcos(theta)=mg, theta will equal 901.1457...degrees, which is approx. 88 degrees.



#14
Sep2304, 07:08 PM

P: 213

Sorry, made a mistake. Doc's method is right, just like my original method.
196 is the tension in each side of the cord. 


#15
Sep2304, 07:08 PM

P: 34

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



#16
Sep2304, 07:09 PM

P: 34

Alright thanks again!



#17
Sep2304, 07:11 PM

Mentor
P: 41,475




#18
Sep2304, 07:13 PM

P: 34

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?



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