how to set this up?

[yoscar04]

No

The component of the "force" along the x-axis will be the tension along the supporting wire (minus any frictional force in the boots). The first line of the problem in part (b) tells you that.

I see, I got it wrong. I confused the question with something else, the tension.

 

[allegansveritatem]

Look at the sketch in response #11. It is a free-body-diagram of your problem.

Use that - to solve for Θ and the component of the astronaut's weight along the y-axis. That is the simulated "weight".

I think I have it right now. I will post my results below. Thanks for the tip.

 

[allegansveritatem]

I went back at it today and I think I finally made some headway. The breakthrough point came while I was working out step b. I had skipped over a for the time being, hoping that something in b would clue me to what I had to do to get a. I worked out the first part of b, for the Moon, I think, and got 80.28 degrees. This bothered me because I knew that my parallelogram was a rectangle and somehow I had it as axiomatic that the diagonal for a rectangle formed 45 degree angles always. So....I decided to find out if that is really true, because my results said it wasn't. I drew a rectangle and put in a diagonal. By eyeball the angles made by the diagonal looked equal. But my diagram was such that there was only a 1 to 2 ratio between the sides. So I tried a 10 to 1 ratio and lo! I saw it! NO No NO! With one exception, the diagonal of a rectangle does not form 45 degree angles...EVER! Happy me! The rest was easy. Here is what I did and I am not saying it is 100% correct but I think it is OK:

 

[Dr.Peterson]

You've got the right answers; did you notice how the various angles in your figure are related? That would help you use theta alone.

Here is a version of the picture that should make things clearer:

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Can you see why the two angles marked theta are equal? I think that's the hardest part of the problem, but it's worth getting used to.​

 

[allegansveritatem]

You've got the right answers; did you notice how the various angles in your figure are related? That would help you use theta alone.

Here is a version of the picture that should make things clearer:

View attachment 20367​

​

Can you see why the two angles marked theta are equal? I think that's the hardest part of the problem, but it's worth getting used to.​

I see that those two angles are equal but I can't say why it should be necessarily so. The angle between the vectors is the same as the angle between the resultant and the x axis is that the expression of some general law?

 

[Dr.Peterson]

One way to see it is that the bottom angle is between the green and horizontal, while the upper angle is between blue (perpendicular to green) and vertical (perpendicular to horizontal). That is, both angles have been rotated 90 degrees, and so stay the same.

Another way is that the bottom angle inside the triangle is the complement of theta, and the top angle is the complement of that, which is therefore theta.

 

[allegansveritatem]

One way to see it is that the bottom angle is between the green and horizontal, while the upper angle is between blue (perpendicular to green) and vertical (perpendicular to horizontal). That is, both angles have been rotated 90 degrees, and so stay the same.

Another way is that the bottom angle inside the triangle is the complement of theta, and the top angle is the complement of that, which is therefore theta.

That is one of the things you can learn from the study of mathematics: The value of knowing even one fact about any given situation. It reminds me of Pope's famous maxim: A little knowledge is a dangerous thing--of course Pope wouldn't recognize his intent in my spin, but my spin nevertheless applies.