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a)$F_{T_1}=3.73N$
b)$F_{T_2}=9.6N$

Giancoli 6th Edition, Chapter 5, Problem 7

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Chapter 5, Problem 7 is solved.

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Transcript for this Giancoli solution

So we have a ball being revolved in a vertical path and in part a we’ll consider when it’s at the top of its path. Let’s draw a picture to illustrate the forces that are on it at this position. So we have the ball there and one of the forces will be gravity straight down, ‘mg’ and we’ll have tension force going the same direction also pulling down. Newton’s second law equation for this situation: the centripetal force ‘Fc’ is equal to ‘mg’ plus the force of tension ‘FT1’ which equal to the centripetal force ‘m’ times ‘v2’ over ‘r’ and rearranging this to solve for ‘FT1’: ‘FT1’ equals ‘m’ times ‘v2’ over ‘r’ minus ‘mg’. Factoring out ‘m’ we have ‘v2’ over ‘r’ minus ‘g’ and we’ll substitute in numbers to get our answer for part a. ‘FT1’ equals zero point three zero zero kilograms times four meters per second all squared divided by the radius zero point seven two meters and we have three point seven three newtons as our answer for part a. For part b we have a slightly different situation, let’s draw a picture of that situation. We’re at the bottom of the arc, here’s the ball, we have the tension force going up ‘FT2’ that's going to be a big long arrow, gravity is still going down and we know gravity is going to be shorter than the tension force arrow because it’s going to be accelerating upwards, there's a net force up, and our Newton’s second law expression will be: ‘FT2’ minus ‘mg’ equals ‘m’ times ‘v2’ over ‘r’. Solving for the tension force: ‘FT2’ equals ‘m’ times ‘v2’ over ‘r’ plus ‘mg’ and then factoring out ‘m’ is ‘v2’ over ‘r’ plus ‘g’ and substituting in numbers we have zero point three zero zero kilograms times four point zero meters per second squared divide by zero point seven two meters plus nine point eight giving us nine point six newtons. We expected a greater force at the bottom than at the top. In the 5th Edition we have a different speed and a different radius of the path so four point one five meters per second instead of four point zero and instead of zero point seven two you have zero point eight five. This changes your answer to three point one four newtons and at the bottom the same changes the answer to nine point zero newtons.

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