no matter how tightly the line is stretched it always sags a little at the center. why?
2006-08-06 00:25:13 · 10 answers · asked by charyz 1 in Science & Mathematics ➔ Physics
Basically, there are three forces at work. One line of force pulls horizontally at each end of the line. The other is gravity, pulling downward vertically.
Imagine that you allow the clothesline to sag visibly. In order for the system to be at equalibrium, all forces must balance out. That means that the horizontal force on each side of the line is equal in opposite directions. It also means that the vertical vector of force on each side of the line (representing line tension) must be equal to the force of gravity acting on the shirt.
Assume that the shirt weighs 2 Kg (heavy shirt. it must still be wet). That means that it exerts a downward force of 2 N (Newtons). Each side of the clothesline must pull upwards with a force of 1 N, to keep equalibrium.
Assume that you've let the clothesline sag, so that it now makes a "V" with an interior angle of 120 degrees. That means that each "leg" of the "V" makes a 30 degree angle with the horizontal plane. Imagine a line at the bottom of the "V", parallel to the ground. This line represents the force pulling horizontally on each side of the line. Imagine another line perpendicular to this one. This vertical line represents the force of gravity. If you place these two lines so that they touch the ends of one "leg" of your "V", you have made a right triangle with half of the clothesline as its hypotenuse. Since we know the amount of force exerted by gravity and we know the angle between the clotheline and the horizontal line, we can apply a little trigonometry to determine the amount of force on one side of the clothesline.
The formula is:
sine (angle of line from horizontal) = (downward force) / (force on the clotheline)
Let's simplify that to:
sine(A) = Fy / Fc
OR
Fc = Fy / sine(A)
So the force on the clothesline is:
Fc = 1N / sine(30)
sine(30) = 0.5
Fc= 1N / 0.5 = 2N
So the force on each side of the line is 2 Newtons. Let's see what happens if we tighten the line a little, so that the line's angle to horizontal is only 15 degrees.
Fc = 1N / sine(15)
sine(15) = 0.259
Fc= 1 / 0.259 = 3.86N
You see that the amount of force on the line increases as the angle becomes smaller. What if we stretch the line super tight, so that the angle is only one-fourth of a degree?
Fc = 1N / sine(0.25)
sine(0.25) = 0.0044
Fc= 1 / 0.0044 = 227N
Our wet T-shirt is exerting the same amount of force as the force exerted by gravity on a five-hundred pound weight! I don't think our little clothesline could take it. The point is, the angle can keep getting infinitely smaller. The pressure on the line will keep getting infinitely bigger as it does. The angle can never reach zero, because doing so would force us to divide by zero, something that we know is impossible.
Hope that helped.
2006-08-06 01:35:30 · answer #1 · answered by marbledog 6 · 1⤊ 0⤋
Shirt is hung at the center means a force equal to mass times “g’ acts vertically down ward.
If the shirt is to be at rest (at equilibrium) there must be an equal but opposite force. That force is called equilibrant.
The equilibrant force has to be provided only by the tension of the rope as there is no other thing which can provide this force.
If the center of the tightly stretched rope is not descending down a little, the rope is horizontal. Horizontal force cannot provide a component in the vertical direction. That means there is no equilibrant. The shirt should come down due to the downward force. Hence it descends a little and the rope inclines a little and now it provides a vertical component enough to balance the shirt.
A rope can be stretched to a maximum tension called elastic limit.
Suppose we have stretched the rope up to that limit. Now if we hung even a small weight the rope will descend a little increasing its tension and the rope will break down.
2006-08-06 01:30:48 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
The force supporting the line is perpendicular to the force (of the shirt plus the weight of the line itself) at the center towards the ground. Forces at right angel do not cancel out each other. Cosine 90 = 0
2006-08-06 03:10:11 · answer #3 · answered by ET 3 · 0⤊ 0⤋
Even without the shirt, the line will always sag. Others have already explained why better than I can.
2006-08-06 04:36:58 · answer #4 · answered by STEVEN F 7 · 0⤊ 1⤋
It has something to do with force vectors. The force vector of the mass of the cloths pointing down doesn't have to be great to overcome the force vector of both sides of the line pointing outwards. I can't remember the full details though.
2006-08-06 00:31:23 · answer #5 · answered by Anonymous · 0⤊ 0⤋
it's due gravity of earth at center of mass specially more at center so if alittle weight is added it goes down.
2006-08-06 00:32:45 · answer #6 · answered by brightstar 2 · 0⤊ 0⤋
from the weight of the line itself.
2006-08-06 00:29:35 · answer #7 · answered by thecat 2 · 0⤊ 0⤋
it is because of the weight of the shirt pulling the clothesline down and the clothesline is elastic
2006-08-06 01:24:22 · answer #8 · answered by raj 7 · 0⤊ 0⤋
could the clothes line be streching
2006-08-06 02:24:38 · answer #9 · answered by jewingengleman 4 · 0⤊ 0⤋
It is called gravity.
2006-08-06 00:30:26 · answer #10 · answered by Anonymous · 0⤊ 0⤋