1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling at an angle

  1. Feb 27, 2014 #1

    adjacent

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=67071&stc=1&d=1393492344.png
    Let the angle of the block(slope) to be 30°.
    height=0.6m, length=1m. Breadth=1m
    Smallest height on diagram is 0m

    Let the red line to divide the block by 2(when viewed from the top)
    The two wheels are of equal diameter and is connected by a rod.
    The total mass of two wheels + rod is 1kg.
    Let the angle be measured from the red line.(At the image,the wheel system is 90° to the line)
    1-Rotate the wheel to 0°.
    It will not move.
    2-Rotate to 90°
    it will fall at maximum speed.
    3- Rotate to 45°
    will be slower than 90°.

    So I am thinking of the relation between the angle and the speed of fall.

    2. Relevant equations
    ##F=ma##
    ....


    3. The attempt at a solution
    Sorry I am at Gr.10. So I haven't studied vectors,forces or speeds at angles.
    -Just curious-
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2014 #2
    A useful property here is the conservation of energy. When the system moves on the incline, its centre of mass gets lower. That means its potential energy is reduced. Which in turn means its kinetic energy is increased. Clearly when it goes "sideways" it needs to go a longer distance to gain the same amount of energy as compared with the case when it goes "straight".

    Alternatively, you can look at the system so that your line of sight is always parallel to the axis of the system. Then you will see that it effectively rolls on an incline with some angle that depends on the "real" incline angle and "sideways" angle.
     
  4. Feb 27, 2014 #3

    adjacent

    User Avatar
    Gold Member

    Thank you.
    Any specific equations for these kinds of "incline" things?
     
  5. Feb 27, 2014 #4
    I think you could try figuring out the "effective" incline angle. This is pure geometry.
     
  6. Feb 27, 2014 #5

    adjacent

    User Avatar
    Gold Member

    I don't understand what you mean by the axis of the system.
    Anyway,I want to find the relationship.
    So I have to make an equation relating incline angle and wheel angle.
     
  7. Feb 27, 2014 #6
    Let the top point of the red line be A. The lowest point of the red line is B. The point on the ground under A is C, so ABC is a right triangle, and the angle ABC is ##\alpha##.

    The system starts at A and rolls down the incline at some angle with the red line and it touches the ground at D. The angle DAB is the "sideways" angle ##\beta##. The "effective" incline angle is ADC. Express angle ADC in terms of ##\alpha## and ##\beta##.
     
  8. Feb 27, 2014 #7

    adjacent

    User Avatar
    Gold Member

    Isn't D the same as B?I said in the OP that lowest point of the red line and the ground has a distance of 0m
     
  9. Feb 27, 2014 #8
    How? The system rolls away from the red line, so it cannot end up at B.

    Do you think I indicated otherwise?
     
  10. Feb 27, 2014 #9

    adjacent

    User Avatar
    Gold Member

    Oh sorry.I thought B as the point of ground.
    Ok.Its a point on the red line and D is a point on ground(Where it touches)

    How's that going to relate wheel angle and speed?
     
  11. Feb 27, 2014 #10
    I do not think you understood me. Points B, C, D are all on the ground. Point B in particular is where the red line touches the ground. Point D is where the rolling system touches the ground. These two points can also be said to lie on the edge of the incline.

    I do not know what the "wheel angle" is and I would like you to answer my question about angle ADC first, anyway.
     
  12. Feb 27, 2014 #11

    adjacent

    User Avatar
    Gold Member

    That seems to be a 3d problem.How am I going to do that?I haven't studied that yet.Can you give me a hint?
     
  13. Feb 27, 2014 #12
    You solve a 3D problem by disassembling it into 2D subproblems.

    In this case, you have three triangles: ABC, ABD, and ADC. These are right triangles and they are all related to each other in some way. Triangle ABD and triangle ADC share hypotenuse AD. Triangle ABC and triangle ADC share cathetus AC. This is enough to express angle ADC via angle ABC (##\alpha##) and angle DAB (##\beta##), it takes just a bit of trigonometry.
     
  14. Feb 27, 2014 #13

    adjacent

    User Avatar
    Gold Member

    Is it OK to include other variables like BD,CB and AC?
     
  15. Feb 27, 2014 #14
    There is an answer that only has ##\alpha## and ##\beta##. But we can discuss your approach even if you do not have that answer yet. Show what you have so far.
     
  16. Feb 27, 2014 #15

    adjacent

    User Avatar
    Gold Member

    OK.
    attachment.php?attachmentid=67086&stc=1&d=1393513693.gif

    I've got
    ##\alpha=tan^{-1}(\frac{h}{b})##
    ##\beta=tan^{-1}(\frac{S}{\sqrt{h^2+b^2}})##
    ##ADC=tan^{-1}(\frac{h}{\sqrt{s^2+b^2}})##

    Not sure how to link ##\alpha##,##\beta## and ADC in one equation :confused:

    EDIT:Latex problem.How do I fix that?
     

    Attached Files:

    Last edited: Feb 27, 2014
  17. Feb 27, 2014 #16
    You meant those:
    ##\alpha=tan^{-1}(\frac{h}{b})##
    ##\beta=tan^{-1}(\frac{S}{\sqrt{h^2+b^2}})##
    ##ADC=tan^{-1}(\frac{h}{\sqrt{s^2+b^2}})##

    (one closing curly brace was missing)
     
  18. Feb 27, 2014 #17

    adjacent

    User Avatar
    Gold Member

    Yeah,it's that.
    How do I link those three together? :confused: :confused:
     
  19. Feb 27, 2014 #18
    As I said:
    So, can you express AD via AB and ##\beta##? Then, can you express AB via AC and ##\alpha##? And, finally, find the ratio of AC with AD?
     
  20. Feb 27, 2014 #19

    adjacent

    User Avatar
    Gold Member

    ##AD=\frac{AB}{cos\beta}##

    ##AB=\frac{AC}{sin\alpha}##

    How do I do that?I don't understand what you said.
     
  21. Feb 27, 2014 #20
    You are required to find a certain angle, let's call it ##\gamma##. Is it related in any way with $$AC \over AD$$ ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted