Anonymous at Fri, 28 Jun 2024 20:47:48 UTC No. 16258708

>>16258580
Three forces act on the ball. They compensate each other. Given that we know of two forces that act on the ball, one (A) in direction -x and gravity (G) in direction -y, the supporting string's tension must be equal to (a*sin(beta),a*cos(beta)).
Since a*cos(beta) = G, this means that the string to be cut must exert a force equal to a*sin(beta) = A.

Cut the string, and the ball at position b now only has two forces, instead of three, acting on it. Gravity (G) and the tension that compensates for it.
Since gravity forces the ball towards -y, the supporting string would theoretically have to compensate for it, which would result in (a*sin(beta),a*cos(beta). However, the string can only compensate along the line (sin(beta), cos(beta)), hence the component pendicular to it cannot be compensated for. Therefore, the supporting string's tension is equal to (b*sin(beta),b*cos(beta)), where b^2 = G^2-A^2

Anonymous at Fri, 28 Jun 2024 21:13:16 UTC No. 16258728

>>16258708
wtf ur a genius

Anonymous at Fri, 28 Jun 2024 21:24:27 UTC No. 16258736

Difficult

Anonymous at Fri, 28 Jun 2024 22:43:42 UTC No. 16258845

1

Anonymous at Fri, 28 Jun 2024 23:19:59 UTC No. 16258905

>>16258708
You forgot air resistance faggot
>>16258580
cant calculate without knowing air resistance and gravity and mass+density of balls and strings faggot

Anonymous at Fri, 28 Jun 2024 23:21:02 UTC No. 16258906

>>16258845
It says ratio of tensions not relative tension so the answer is 1:1 not simply 1 faggot

Anonymous at Fri, 28 Jun 2024 23:24:44 UTC No. 16258910

>>16258905
The dimensions of the mass aren't given, you can't estimate air resistance. Besides, the ratio of forces will be the same, only the point B would move.

Anonymous at Fri, 28 Jun 2024 23:49:02 UTC No. 16258955

>>16258580
This is a basic draw a picture problem. OP should've posted this on stupid questions if you need homework help. You killed a thread for this. If you need a hint for the second tension, tilt your head. When simplified, your answer should be related to cos(beta).

Anonymous at Fri, 28 Jun 2024 23:50:53 UTC No. 16258961

>>16258955
also, notice that when Beta equals pi/2, the ratio should be "infinite" because when initially taut, the horizontal tension force should be "infinite" to keep the ball at that top height. Obviously irl it droops a little, just like irl the rope also has mass.

Anonymous at Sat, 29 Jun 2024 00:58:58 UTC No. 16259081

>>16258955
It's not homework. Stop trying to get me banned. I just feel like these problems bring up some interesting discussion

Anonymous at Sat, 29 Jun 2024 03:16:37 UTC No. 16259221

>>16258708
>he believes in gravity

Anonymous at Sat, 29 Jun 2024 03:21:48 UTC No. 16259228

>>16258580
Why is this a three dot problem? Consider only the force in the vertical direction.
A: Tsinb = mg
B: Tsinb = mg
Divide the two equations. T_a = T_b

Anonymous at Sat, 29 Jun 2024 03:49:13 UTC No. 16259261

>>16258906
1/1 = 1
learn math and mannors

Anonymous at Sat, 29 Jun 2024 07:15:58 UTC No. 16259426

1/cos(beta)

Anonymous at Sat, 29 Jun 2024 09:05:55 UTC No. 16259499

>>16259081
It's not interesting, it's just a first year FBD problem.

Anonymous at Sat, 29 Jun 2024 09:14:45 UTC No. 16259504

>>16259499
I feel like this exercise would make a perfect introduction into Hamiltonian/Lagrangian mechanics. Knowing that the ball is constrained and can only follow a path f(x,y,z) = 0 would allow you to transform the exercise into one involving two forces.

Anonymous at Sat, 29 Jun 2024 11:21:59 UTC No. 16259587

>>16259499
Ok? Not all of us have 120IQ and can breeze through this.

Anonymous at Sat, 29 Jun 2024 11:28:48 UTC No. 16259592

>>16259504
it's not that deep lil bro

Anonymous at Sat, 29 Jun 2024 14:52:07 UTC No. 16259861

>>16258580
Can uou not solve this by just assuming ball has mass m and drawing a force diagram. The x component of the string at the top would be the tension on the other string.