how to set this up?

[allegansveritatem]

Here is the problem:



I am interested in part a for now. Here is the best (relative to my other attempts, so you can imagine what they must be like) of a number of attempts I made:


Obviously I am groping in darkness here but...how to think about this problem? I now see this much, that 160 is not what I have here but is one of the components of a set of vectors whose combination will give me my solution. No?

 

[yoscar04]

Hi, from where did you get that the angle [MATH]\theta[/MATH] is 45 degrees?

 

[yoscar04]

I think you will need for your calculation g(Moon)=1.625 m/sec², and g(Mars)=3.7 m/sec². Calculate the components of the weight along the x-axis and you will be done. Review the definition of cos and sin for the angle [MATH]\theta[/MATH] in your draw.

 

[Dr.Peterson]

Your work looks fine except that you didn't leave [MATH]\theta[/MATH] as a variable, and replaced it with a specific value, 45 degrees.

I suspect that you didn't realize that the answer to (a) will be an expression; you will use that expression in (b), where you will set y to 27 or 60 and solve for [MATH]\theta[/MATH].

 

[allegansveritatem]

Your work looks fine except that you didn't leave [MATH]\theta[/MATH] as a variable, and replaced it with a specific value, 45 degrees.

I suspect that you didn't realize that the answer to (a) will be an expression; you will use that expression in (b), where you will set y to 27 or 60 and solve for [MATH]\theta[/MATH].

Thanks for response I have not been able to get online since 10 last night so I have not seen this til now. I will reply later this evening after I have had a chance to digest your post.

 

[allegansveritatem]

Hi, from where did you get that the angle [MATH]\theta[/MATH] is 45 degrees?

thanks for response. My internet carrier seems to have used the last 16 hours or so to revamp the system and meanwhile all us poor customers have not had any internet. I will reply to your post later this evening after I have had a chance to study it and the other posts in the thread.

 

[allegansveritatem]

Hi, from where did you get that the angle [MATH]\theta[/MATH] is 45 degrees?

I think the 45 came from the fact that the angle that the spaceman makes with the inclined plane is a right angle--or from the idea that the parallelogram that I imagined was actually a rectangle, with the long sides being the rope and the short sides the spaceman. The diagonal of a rectangle provides the 45 degree angles.

 

[allegansveritatem]

I think the 45 came from the fact that the angle that the spaceman makes with the inclined plane is a right angle--or from the idea that the parallelogram that I imagined was actually a rectangle, with the long sides being the rope and the short sides the spaceman. The diagonal of a rectangle provides the 45 degree angles. I know that in my diagram I have a square and I'm no longer sure how I came up with that.

 

[allegansveritatem]

I think you will need for your calculation g(Moon)=1.625 m/sec², and g(Mars)=3.7 m/sec². Calculate the components of the weight along the x-axis and you will be done. Review the definition of cos and sin for the angle [MATH]\theta[/MATH] in your draw.

I will try that but first I have to get part a straight.

 

[allegansveritatem]

Your work looks fine except that you didn't leave [MATH]\theta[/MATH] as a variable, and replaced it with a specific value, 45 degrees.

I suspect that you didn't realize that the answer to (a) will be an expression; you will use that expression in (b), where you will set y to 27 or 60 and solve for [MATH]\theta[/MATH].

I was being guided by the fact that there seems to be right angle in this problem and wouldn't that indicate that the parallelogram method would yield a rectangle and a diagonal across a rectangle would give us 45 degree angles? I will have to work on this again tomorrow. You are right,. I was shooting for a specific answer. I am planning on go over the chapter on vectors again from the start because I am still not clear what is going on in these problems. I may be trying to apply things that don't need to be applied. I will post my results.

 

[yoscar04]

Hi, I think you have to consider the angle [MATH]\theta [/MATH] given in the problem. They ask you to find the angle [MATH]\theta[/MATH] appropriated to simulate the Moon and Mars gravity conditions. Mark the angle theta as it is shown in your book picture and write the components of the weight along the x and y axis. Check that the angle between the y axis and the weight is also [MATH]\theta[/MATH]. Then you will see that W_x=mgsin[MATH]\theta[/MATH], and W_y=mgcos[MATH]\theta[/MATH]. Please see draw attached. Remember that the angle between the axis x and y is 90 degrees. You may even check that your answers are right by considering the particular cases 0 and 90 degrees.

 

[allegansveritatem]

Hi, I think you have to consider the angle [MATH]\theta [/MATH] given in the problem. They ask you to find the angle [MATH]\theta[/MATH] appropriated to simulate the Moon and Mars gravity conditions. Mark the angle theta as it is shown in your book picture and write the components of the weight along the x and y axis. Check that the angle between the y axis and the weight is also [MATH]\theta[/MATH]. Then you will see that W_x=mgsin[MATH]\theta[/MATH], and W_y=mgcos[MATH]\theta[/MATH]. Please see draw attached. Remember that the angle between the axis x and y is 90 degrees. You may even check that your answers are right by considering the particular cases 0 and 90 degrees.

I worked on this again today after reviewing the chapter on vectors. I think that I did something like what you are suggesting...but I went somewhat astray in the direction of a baroque elaboration--too baroque to fix I'm afraid. I did get some true expressions for X and Y but...well I didn't even use them when I did part b. Here is what I got for a:

and here is what I got for part b:

 

[Dr.Peterson]

Your picture here has nothing to do with the problem, and you've brought in angles alpha and beta, which are not there.

Please start over for part (a). And don't use the law of sines or anything; you just need right triangles. Your original drawing, the third in post #1, was fine except that what you called 45 should have been called theta.

 

[yoscar04]

I worked on this again today after reviewing the chapter on vectors. I think that I did something like what you are suggesting...but I went somewhat astray in the direction of a baroque elaboration--too baroque to fix I'm afraid. I did get some true expressions for X and Y but...well I didn't even use them when I did part b. Here is what I got for a:
View attachment 20344
and here is what I got for part b:View attachment 20345

Did you understand my draw?

 

[allegansveritatem]

Your picture here has nothing to do with the problem, and you've brought in angles alpha and beta, which are not there.

Please start over for part (a). And don't use the law of sines or anything; you just need right triangles. Your original drawing, the third in post #1, was fine except that what you called 45 should have been called theta.

Here is the crux and the hinge point of my predicament: How to transfer the info of the diagram to a Cartesian plane. You are right, When I drew it the first time I went on the right angle assumption and when I did part b (for better or for worse) I switched back to that assumption. I will give it another go today.

 

[allegansveritatem]

Did you understand my draw?

well, as the feller said,"t seemed like a good idea at the time". Truth is I didn't quite know how to transfer the information to an xy plane. But what I am going to do now is assume that there is a right angle in the mix and go from there.

 

[yoscar04]

well, as the feller said,"t seemed like a good idea at the time". Truth is I didn't quite know how to transfer the information to an xy plane. But what I am going to do now is assume that there is a right angle in the mix and go from there.

When you project a vector to the axis x or y, it forms a 90 degrees angle with each axis.

 

[Deleted member 4993]

well, as the feller said,"t seemed like a good idea at the time". Truth is I didn't quite know how to transfer the information to an xy plane. But what I am going to do now is assume that there is a right angle in the mix and go from there.

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".

 

[yoscar04]

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 you meant the weight along the x axis, right?

 

[Deleted member 4993]

I think you meant the weight along the x axis, right?

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.