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Critical angle (torque) at which the block tips over

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    I know this tipping a block question has been asked many times and I know how to find the critical angle at which a block tips over.

    I was just wondering how one would go about analyzing the torque of the system. We all know that anywhere from 0 to the critical angle, the block wont tip over around the left bottom vertex. I wanted to show the block wont rotate by showing the net torque of the system is zero. I can easily show it for zero angle slope but for some arbitrary angle, it doesn't go to zero...
    It makes intuitive sense that the block wont rotate because the block wont wade into the slope...
    but I want to see the net torque going to zero..

    (just to let you guys know, I know that at the critical angle, the only source of torque is the gravitational weight because the normal force and the friction all pivot at the left end of the base, so r = 0)


    2. Relevant equations

    Torque = r x F

    3. The attempt at a solution
    (there is no friction)

    I have attached my work.

    So, according to my diagram, at the critical angle, Alpha = Theta. when the CG is right above the left end of the base, in which case, the block starts tipping over.

    I guess my work is valid only when Theta =0 (when the block is just resting on a horizontal surface, in which case Fn= Fg and r x (Fn + Fg) = 0, which I can show algebraically )
     

    Attached Files:

    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 20, 2014 #2

    Simon Bridge

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    OK - the situation is that you are tipping the surface and you want to know when the surface is too steep for the block to fall over.

    I notice you left off the friction force ... anyway, the normal force depends on gravity and the angle, and it is only just big enough to stop the block falling into the ramp. In terms of torques: the normal force supplies a counter-torque to that supplied by the weight ... until the critical angle is reached. After the critical angle, there is no normal force.

    The angle ##\small \alpha## can also be expressed in terms of a and b (height and width) along with ##\small \theta##.

    That and resolving the forces wrt ##\small{\vec{r}}## should get you where you need to go.
     
    Last edited: Jan 20, 2014
  4. Jan 20, 2014 #3
    thank you for your reply.

    The question is set up such that friction doesnt exist.

    Also, i took into account the fact that Fn = Fg cos (theta)

    i dont see what part of my work is wrong because my conclusion for some arbitrary angle less than the critical angle, (when the base is fully in contact with the ramp that is) is that there is non zero net torque.

    my guess is, for some angle less than the critical angle, the normal force is (for the purpose of approximation) the normal force is evenly distributed over the base surface and you have to integrate the contribution over the base, (can no longer approximate it as Fn and Fg acting on the CG) and the net torque comes out to be zero.
     
  5. Jan 21, 2014 #4

    Simon Bridge

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    If you have a net torque for sub-critical angles, then you either failed to account for all the forces or you have not properly included the reaction force.

    I suspect you may be using the angle theta to represent two different things.


    Try: define ##\phi## = angle of ##\vec r## to the ramp.
    thus the angle of ##\vec r## to the horizontal is ##\theta+\phi##.

    an unbalanced torque about the corner results in an angular acceleration of the box:

    ##\sum \tau = I\ddot{\phi}##
     
  6. Jan 21, 2014 #5
    i have defined alpha and theta, so your phi is 90 - alpha.

    I feel like for sub critical angles, this CM treatment in calculating the net torque may not work.

    Just like when it is at the critical angle, Fnormal and Ffrictin do not come into play because they are drawn from the pivot point, not the CM
     
  7. Jan 21, 2014 #6

    Simon Bridge

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    So: ##\sum\tau = -I\ddot\alpha## ... is that what you did?

    That's right - only one component of each force goes through the center of mass, the sum of these contributes to translational motion - the other component is perpendicular to the moment arm and contributes the torque.

    Below the critical angle all the torques cancel out, above the critical angle the torques become unbalanced - because the direction of the gravity torque has changed.
     
  8. Jan 21, 2014 #7
    Yes, and the net torque according to my work for sub critical angles does not absolutey go to zero... and I have not been able to identify what i did wrong...

    the only explanation to my net Torque not going to zero is for sub critical angle where the flat base is fully in contact with the slope, it may not make sense to simplify the picture by saying all the forces can be anchored at the CM...

    Thank you so much for getting back to this thread...
     
  9. Jan 22, 2014 #8

    Simon Bridge

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    OK - I'll start you off:
    For sub-critical angles: ##\phi## is a constant.
    1. Torques about the front corner:
    ##\qquad \frac{1}{4}mg\sqrt{a^2+b^2}\cos(\theta+\phi) + \tau_N = 0## ... because: not rotating.

    2. Torques about the back corner: ...

    etc.

    note: ##b=\sqrt{a^2+b^2}\cos\phi ,\; a=\sqrt{a^2+b^2}\sin\phi##

    aside: does it matter that the reference frame is non-inertial?
     
  10. Jan 22, 2014 #9
    my work based on your angle definition

    well.. i have calculated the net torque based on the way you defined theta and phi...

    i dont see how the general expression goes to zero...
     

    Attached Files:

  11. Jan 22, 2014 #10

    haruspex

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    You've assumed the normal force always acts through the middle of the base. In general, the supporting force is distributed across the base in an unknown way. (By assuming a slight uniform springiness, you can suppose that it is distributed in a manner that increases linearly from one corner to the other.) But what matters is that the net supporting force can act anywhere within the base, as required to make the net torque zero. The block will topple when that is not possible - i.e. when even if all the supporting force acts through the lower corner it still cannot provide enough counterbalancing torque.
     
  12. Jan 22, 2014 #11
    Ok.. thats what i wanted to hear..

    I wanted to get confirmation from someone that simplifying the torque analysis by saying all forces act through the CM is no longer valid in this case.

    Thank you
     
  13. Jan 23, 2014 #12

    Simon Bridge

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    @hangainlover: this is what I was hoping you'd see by doing the torque analysis first and then look for where the action of the forces must be. This would give you an effective moment arm for support and we could have talked about how this comes about. Haruspex beat me to it.

    You've probably converted distributed forces into effective point forces before - like for a static beam.
    A cute variation on your problem is to put the block on two point supports and work out the forces... i.e how does the critical angle depend on the position of the supports (if at all)?
     
  14. Jan 23, 2014 #13

    haruspex

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    But that isn't what you did. You took the normal force as acting through the centre of mass (and hence through the centre of the base), but you took the frictional force as acting along the base. Had you taken friction as also acting through the centre of mass it would have been ok.
    More precisely, though, all three forces act through the point on the base that's directly below the centre of mass.
     
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