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Maximum acceleration before tipping occurs

  1. Jan 9, 2009 #1
    Folks, I'm a newbie. I was given this problem to illustrate the concept of accounting for inertia in sliding friction or centripetal force problems, by a tutor. The tutor is unavailable and I am having trouble with the resolving the problem. The tutor's method to solve the problem was by equilibrium of moments, using simultaneous equations. I tried to find a trigonometry solution. I have come up with two different answers. I have an answer given by the tutor and was using it to work backwards, for my trigonometry solution. I think I must be missing a principle.

    I did a search here and found similar questions but nothing quite the same. The difference being this box is already sliding, so it's not a question of sliding vs tipping. It's a question of how fast the box can be slid before tipping commences.

    A box has a mass of 45 kg and is dragged along the ground with an acceleration by the horizontal force P. The centre of gravity of the box is 0.9 m above the ground and the base is 1.25 m wide. The force acts 1.2 m above the ground. The co-efficient of friction is 0.2. Calculate the maximum acceleration before tipping occurs. Answer given was 12.55 m/s2


    I've scanned my work into .pdf format. I'll attached three documents. The first is my attempt at the trig solution, second is my working of the moments solution, the third is my notes from the tutorial.

    Hope this is viewable.

    Cheers,

    oldmancan
     

    Attached Files:

    Last edited: Jan 9, 2009
  2. jcsd
  3. Jan 9, 2009 #2
    In my diagram the pulling force P acts 1.2m above the bottom leading corner.
    The weight acts through the middle of the base 0.6125m behind the leading bottom corner.
    Taking moments about the bottom leading corner -
    Force P accelerates the mass with acceleration A, and overcomes the friction which is 45 times 0.2 times G newtons. P = 45 x A + 45 x 0.2 x 10 or 45A + 90
    Taking moments about the bottom corner:

    P x 1.2 = 45 x 10 x 0.6125 Newton Metres - When clockwise moment equals anticlockwise moment about the pivot and the box is at the point of tipping.

    1.2 x (45A+90) = 275.625

    My answer is A = 3.104 metres per second squared.

    However, this is an oversimplification. The accelerating force always tends to rotate the box and this concentrates the floor's reaction towards the leading edge. We can only assume that at all times, the total friction force is always 0.2 times mG. irrespective of whether the box is about to flip up onto its leading corner or is still evenly resting on the floor. Fortunately, the friction force is always located at the base and always directed straight through the leading corner, so its distribution over the base is irrelevant.

    On Second Thoughts:
    We need look at the pulling force P in two parts. The part which overcomes the friction is 45 x 10 x 0.2 Newtons and it causes a clockwork moment about leading corner of 90 x 1.2 Nm.
    But the part of P which is accelerating the box is working against the box's inertia which is at its c of g. hence the clockwise moment of this part is only 45 x A x 0.3 Nm. The 0.3 is P's distance above the c of g. Hence that total clockwise moment is 90 x 1.2 + A x 45 x 0.3 Nm.
    The anticlockwise moment about the leading corner is the same : 45 x 10 x 0.6125 Nm.
    Equating these and solving for A, I now get A = 12.42 ms^(-2)

    Sorry I was wrong first time.

    It helps to imagine what would happen if the pull was applied at the c of g. The pulling force would not have any contribution to the tipping moment. As the friction and the weight are both constant (in theory) then the box would tip over before it started moving or it would never tip over irrespective of speed or acceleration.

    I hope that's right.

    Lastly it seems to me that the middle of the base is 0.6125m from the corner, not 0.625 as some solutions assume.
     
    Last edited: Jan 10, 2009
  4. Jan 9, 2009 #3
    In my diagram the pulling force P acts 1.2m above the bottom leading corner.
    The weight acts through the middle of the base 0.6125m behind the leading bottom corner.
    Taking moments about the bottom leading corner -
    Force P accelerates the mass with acceleration A, and overcomes the friction which is 45 times 0.2 times G newtons. P = 45 x A + 45 x 0.2 x 10 or 45A + 90
    Taking moments about the bottom corner:

    P x 1.2 = 45 x 10 x 0.6125 Newton Metres - When clockwise moment equals anticlockwise moment about the pivot and the box is at the point of tipping.

    1.2 x (45A+90) = 275.625

    My answer is A = 3.104 metres per second squared.

    However, this is an oversimplification. The accelerating force always tends to rotate the box and this concentrates the floor's reaction towards the leading edge. We can only assume that at all times, the total friction force is always 0.2 times mG. irrespective of whether the box is about to flip up onto its leading corner or is still evenly resting on the floor. Fortunately, the friction force is always located at the base and always directed straight through the leading corner, so its distribution over the base is irrelevant.
     
  5. Jan 9, 2009 #4

    Delphi51

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    The trig solution is a complete mystery to me! I don't see any angles in the problem, so no trig applies.

    The tutor solution looks good.
    The counterclockwise moments about the leading bottom edge are:
    .9ma + .625mg due to the inertial force backward at vertical center and the weight of the object acting downward from the center .625 m back from the leading bottom edge.

    The clockwise moment is the total pulling force (F=ma plus Friction) times 1.2 meters:
    1.2(ma + .2mg)

    Set the moments equal, solve for a = 12.59 m/s2.
    There aren't any simultaneous equations - only the one equation to solve.


    Oh . . . now I'm seeing trigonometry as it begins to tip. The vertical center will rise and the distance to the mg force will decrease as the angle of tip increases. But I still think the trig does not apply because we are only asked for the acceleration when it BEGINS to tip and the angle is zero.
     
  6. Jan 10, 2009 #5
    Delphi51,

    Just because we don't see something doesn't mean it doesn't apply. I believe all force problems can be represented by trigonometry, we just have to understand how. See the following attachments, FR_parallel and FR_inclined. In both, F is the force to overcome friction and start the box moving.

    I suppose I should have wrote simultaneous expressions, not equations. My terminology is lacking. You did the same thing without thinking in the terms I used.

    1st expression: F = Ma + 0.2 x Mg
    2nd expression 1.2 x F = 0.9 x Ma + 0.625 x Mg

    Dchmm,

    I see your edit and still think something is missing. I reworked your solution and where you answered 12.42 m/s^(2), I get 12.64 m/s^(2). The difference being I used 9.81 instead of 10 for g and 0.625m for mid-width of box. 1.25/2 = 0.625.

    All,

    I don't really care what the answer is I want to understand what principles are in effect. Can anyone see what is wrong with my trigonometry solution and explain it?

    My gut feeling is that the horizontal component of P = Fr + Fi + F?
    Fr = friction force
    Fi = inertia force
    F? = tipping force

    Thanks for your thoughts.

    Cheers, omc
     

    Attached Files:

    Last edited: Jan 10, 2009
  7. Jan 11, 2009 #6

    PhanthomJay

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    we haven't seen your latest attachments yet; i hope you expalin what you mean by F_incline. But i notice that your original 'space' and FBD diagrams appear to be incorrect. In the space diagram u have a force R, which, if that represents the normal force, acts , at the point of tipping, straight up at the lower right corner, not through G. Then you identify some force called F, which does not belong there. It's tough enough having a pseudo force F_I , without adding more.
    In your FBD, there are too many forces shown. If you are attempting to show the forces all acting through G, then you have the pseudo force F_I, and the weight force, shown correctly; but get rid of those F and F' values', and R acts vertically up through G, and P acts horizonatlly through G, neither at an angle. Then you have to add the 'couples' at G due to the torques causeded by P, F_r, and R_n (.3P cw, .9F_r cw, and .625R_n ccw. These can be represented as vectors into or out of the page, if you wish.
    Your trig solution eludes me. You need 2 separate diagrams; the first for the forces, which will close to the zero point; the second for the torque vectors in the z direction, which also must close to zero.
    I like your tutor's solution the best.
     
    Last edited: Jan 11, 2009
  8. Jan 13, 2009 #7

    PhanthomJay

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    In your latest set of attachments, you apparently have reworded the problem to ask
    "What is the horizontal force, P, required to just overcome friction and start to accelerate the block when (a) the horizontal force P is applied through the c.g. of the block, and (b) when the horizontal force P is applied 0.3m above the c.g. of the block."
    You then go on to show, in a rather tedious fashion, for understanding, that the force required is the same in both cases. This is a natural consequence of Newton's first law, sum of all forces in the x direction =0, regardless of their position with respect to the c.g. (that is, in either case, P - umg = 0, or P = 88.29 N; however, in general, there is no guarantee that the block won't tip over before it starts to translate under this force, because sum of torques about any point must also sum to zero, to prevent rotation).

    Now, although you have arrived at the correct answer for both cases, your diagrams appear incorrect. For the first case, which you refer to as the 'parallel' case, your space diagram shows a force F. That should be designated as P. Then you show a force R, which apparently is a resulltant of the Normal and friction force? If so, you can't show F_R in your diagram, because it is included in R; also, although its orientation (phi angle) is OK, R does not pass thru G. It is best to show the Normal force, N, as a perpendicular concentrated force acting upward at the base; its distance from the base can be determined by summing moments of all forces = 0 about any point; and F_R, the friction force, acts at, and parallel to, the base. Bottom line: forget about F and R!

    The 'space' diagram can serve as the free body diagram. however, if you wish to show all forces acting thru the c.g, then P acts to the right, F_R to the left, W down, and N up. All forces must add to zero (P = F_R and W = N). But now you must include in your diagram the moments (torques), (known as couples) about the c.g. , as produced by F_R, and N.
    Your vector diagram should then close to zero in both the x-y and x-z planes (The couples can be represented as torque vectors in the z direction).

    Similary, proceed the same way for what you call the 'incline' case. There is no 'incline'. P acts horizontally; get rid of F and R; add N, which acts upward at the base, at a different location than the prior case.
     
    Last edited: Jan 13, 2009
  9. Jan 13, 2009 #8
    Sorry for any confusion. I posted the latest set of attachments for the benefit of Delphi51. He said he couldn't see any angles so there couldn't be a trigonometry solution.

    I am working the solutions the way my text and tutor have been instructing me. I have an exam looming and don't have time to worry about this question. I won't pursue this any further. I'll talk to my tutor in a few months and get it straight with him then.

    All the best to all of you. Thanks for your interest.

    Jay, I'll give thought to your comments. I don't see a problem with breaking P into components. I used horizontal components acting through G, centre of gravity, because the inertia force acts through G.

    cheers, omc
     
  10. Jan 13, 2009 #9

    PhanthomJay

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    I guess with P's, F's, F' 's, R's, F_R's, N's, W's, and 'psuedo' inertia forces, we're not together on the terminology. I fail to see how a horizontal force, P, can have a component. Also, in the latest attachment, there are no inertial pseudo forces. But in any case, best wishes in your exam. With all the effort you've put into this problem, I'm sure you will do well. This problem is a step above the 'introductory' level.
     
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