_{y}= 2Tsin(30) -W = 0

Find the tension in the clothesline in terms of W.

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In this problem, the center of the clothesline is a point in equilibrium, i.e., the sum of all forces equals zero. There are two tension forces of the same magnitude T directed at 30° above the horizontal pointing left and right from the center, as well as the weight W pointing straight down. The vertical or y-component of either tension force is the opposite side in a right triangle whose hypotenuse is T and is therefore given by T sin(30). Then the sum of the forces in the vertical direction equals zero:

∑ F_{y} = 2Tsin(30) -W = 0

Since 2 sin(30) = 2 (1/2) = 1, we have

T-W=0, or T=W.

The tension in the clothesline equals the weight.

The weight is not given as a value. It is merely labeled as "W".

Antoine D. | Experienced and Motivated Science TutorExperienced and Motivated Science Tutor

do you mean 30 degree angles?

i am assuming the clothesline is shaped like a triangle. you divide the triangle in half with a line thru it and now you have 2 right triangles ( 30-60-90). there are two hypotenuses and their tension is found by W/sin30deg. Since there are two hypotenuses, it should be 2 times W/sin30deg; since the tension it supported by the two sides of the clothesline.

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