Physics + tension

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br2pi5

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Having some trouble understanding these questions. I'd appreciate the help!

1) the 3kg of clothes one, what would be the set up?
2) the other question, I set it up as sin45 = Ft/T1, so Ft= T1sin45 which then equals 300N (Fg). Is that right?
 

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Also, shouldn't this be Fg=mgcos(angle) because it's on an incline?
 

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Having some trouble understanding these questions. I'd appreciate the help!

1) the 3kg of clothes one, what would be the set up?
2) the other question, I set it up as sin45 = Ft/T1, so Ft= T1sin45 which then equals 300N (Fg). Is that right?

For the kid on a pendulum question, you have set it up correctly, where T = mgcos(theta) at the apex, where v = 0, so mv2/r = 0.

For the clothesline question, you need to draw a picture. Initially you have a horizontal cable that is 6 ft long. It then sags in the middle to a point that is lowered by 3 ft. So you have a scenario where the tension must be found using a triangle. The triangle has vertical side of 3 ft (the amount it sags). It has a horizontal side of 3 ft (half to the original ft m line). This means it's a 45-45-90 triangle and the hypothenuse represents the tension. However, there are TWO cables holding the mass up, so you need to consider BOTH. Each side is responsible for supporting 15 N vertically. So if you look back at your triangle, the vertical side is 15 N. Because it is a 45-45-90 triangle, the hypotenuse is 15(root 2) N. The total tension, for both lines, is 2 x 15(root2) N = 30(root2) N.
 
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