Some Tough Problems (for me at least)

  • Thread starter tony_n88
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  • #1
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Hi, I'm new here. I came to this board because I am having troubles with certain problems with an assignment. If anyone can help me, it would be greatly appreciated.

http://www.freewebtown.com/qdn745/Physics%201.JPG [Broken]
http://www.freewebtown.com/qdn745/Physics%202.JPG [Broken]
 
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  • #2
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Yes we all help each other on PF...but the first rule is to post all your work here :smile: before asking for help...
 
  • #3
berkeman
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The second rule is to be careful about the size of your pictures. Try to make them fit in a standard widow without the need to scroll horizontally.

Now, how would you go about approaching the first two problems?
 
  • #4
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I just solved the first problem, can someone verify if it is correct (methods, notation wise)?http://www.freewebtown.com/qdn745/ISP#1.JPG [Broken]
 

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  • #5
Hootenanny
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It may take a while for you image to be approved, would you consider typing your solution?
 
  • #6
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Lets try this, I uploaded the image on a free server.
http://www.freewebtown.com/qdn745/ISP1.JPG [Broken]
 
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  • #7
Hootenanny
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I can't see any problems with your workings in question one. What are your thoughts one question two. If you would like to include equations in your posts,https://www.physicsforums.com/showthread.php?t=8997" Would you also please consider resizing your image in you orgininal post.
 
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  • #8
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Thanks. I just fixed it, sorry about that. I'm working on number 2 now, It will take me awhile to understand+solve it.
 
  • #9
Hootenanny
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Thank you, much better. :smile: I'll give you a hint for question two - consider the torques about the top of the step.
 
  • #10
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Here is what I got for number 2, I hope it's right.
http://www.freewebtown.com/qdn745/ISP2.JPG [Broken]
 
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  • #11
Pyrrhus
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Looks good. Nice format. By the way, something interesting about this problem is that the magnitude of the weight and the applied force are the same for the Ry and Rx components of the reaction force at point A.
 
  • #12
Hootenanny
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Again, your working looks spot on and you got the same answer as me :smile: 5 stars for presentation I think
 
  • #13
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Thanks guys. Now for number 3, I'm not sure how to solve this problem but heres my go at it. First I am supposed to find the center of mass of the uniform triangle which is where the force of gravity is concentrated on, from there I would just calculate the torque as usual with the moment arm from the point of rotation to the center of mass and Fg = mg. Is this the correct approach?
 
  • #14
Hootenanny
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tony_n88 said:
Thanks guys. Now for number 3, I'm not sure how to solve this problem but heres my go at it. First I am supposed to find the center of mass of the uniform triangle which is where the force of gravity is concentrated on, from there I would just calculate the torque as usual with the moment arm from the point of rotation to the center of mass and Fg = mg. Is this the correct approach?
Sounds about right. Looks like your getting the answers without our help, we'll be redundant soon :wink:
 
  • #15
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Here is what I did for number 3,
http://www.freewebtown.com/qdn745/ISP3.JPG [Broken]

I'm not very certain about my statements (especially for part b) :shy:
 
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  • #16
Hootenanny
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(a) Is spot on.

For (b) you may also want to add that "there is no horizontal displacement betweent he centre of gravity and the axis of rotation." But your answer is technically correct.

(c) Also looks correct to me although you may want to state that assuming AC = 1/2AB.
 
  • #17
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Hootenanny said:
(a) Is spot on.

For (b) you may also want to add that "there is no horizontal displacement betweent he centre of gravity and the axis of rotation." But your answer is technically correct.

(c) Also looks correct to me although you may want to state that assuming AC = 1/2AB.
Okay, it's now clear to me, thanks again for your help. Two more questions to go....

For the next question, I'm completely new to these types of problems. I'm having troubles actually just understanding the problem, but I drew a diagram of what i think is going on,
http://www.freewebtown.com/qdn745/ISP4A.JPG [Broken]
The blue arrows represent the forces.
 
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  • #18
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Once again, here is my uncertain solution,
http://www.freewebtown.com/qdn745/ISP4B.JPG [Broken]
I've got a real bad feeling about this one.
 
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  • #19
Pyrrhus
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Each roof rafter supports a 170kg load, in reality snow loads are modeled as distributed, not concentrated. Also they usually calculate the load on a horizontal roof by using the load on the soil, and if the roof has a slope, you apply a factor... How did you know the location of the snow loads on each rafter? did you guess it?. In my opinion i think they will be at L/4 the length of the house.
 
  • #20
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Yeah I know I'm totally new to this question, I don't know how to proceed in the method that your explaining. If you can guide me thorugh this question, that would help a lot, because I'm totally lost.
 
  • #21
Pyrrhus
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What was you idea?


The snow load should be vertical (that's how it is), maybe what they meant about contact forces at the peak (a force with unknown direction, 2 components), beak apart the rafters from each other, you will have their snow load, a unknown force at the peak (neglecting internal moment) and the vertical force from the wall junction (also neglecting internal moment and horizontal force). i forgot and of course Tie beam force with unknown direction, 2 components.
 
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  • #22
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I don't know, as you said, I guessed my concepts for this question that the force is concentrated and that it divides evenly into two components. I'm stumped with this question.
 
  • #23
Pyrrhus
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tony_n88 said:
I don't know, as you said, I guessed my concepts for this question that the force is concentrated and that it divides evenly into two components. I'm stumped with this question.
You mean the snow load??, you decomposed it in 2 components, and your picture shows a y-axis parallel to the force, so it wouldn't have 2 components. No?
 
  • #24
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Your right and I thought it would create a force perpendicular to the slant... If anyone knows how to properly approach this question, any help would be appreciated.
 
  • #25
Pyrrhus
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This is what it looks like to me:

1) The snow load doesn't act perpendicular to the rafter, this is because it's modelled like an uniform distributed load perpendicular to the x-axis for engineering purposes (ASCE 7 Standard).

2) The Rafter's joint, has 1 force with unknown direction, aka 2 unknown components. In reality there would also be a internal moment, because the rafters are rigidly joined, at least it shouldn't rotate about an axis coming out of the graphic. (Note: the book talks about contact forces concentrated at the roof peak)

3) The walls and rafters are rigidly joined, too aka internal moment and internal force with unknown direction. However the book asks us to neglect the horizontal component, also we will be neglecting the moment.

4) The point of conection (articulation) between each rafter to the Tie Beam has a force with an unknown direction (2 components).

I think the way to tackle this problem should be like a frame. Imagine you take out the walls and only leave the upper part. You will have 2 unknown normal forces at the bottom (Wall connection to the rafter), plus two known forces acting on our frame.
 
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