1. Limited time only! Sign up for a free 30min personal 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!

Potential Energy & Tippipng Forces

  1. Oct 5, 2009 #1
    Hi All,

    I have a problem i am trying to figure out, if an object (scale drawing in attachment) is travelling at a constant velocity of 2.16 m/s, with a mass of 496 Kgs, what is the potential energy required to 'tip' this object over?
    Kinetic energy = 1/2 Mass x Velocity^2=1/2x496x2.16^2 which = 1157.07 Joules, all good, however;
    Potential Energy= Mass x (sqrts^2+a^2-s)=496x(sqrt6.09^2+0.79^2)-6.09 = 496 x (6.14-6.09) which =25.3 Joules
    which means, at this velocity this object will tip if it strikes an object on the ground.
    Does this seem correct?
    Also, how can i calculate the force required to tip this object over?
    I have put scale drawing in as an attachment to help.
    I am confused and at a complete standstill.

    Any help is greatly appareciated.
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2009 #2

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    glocki35: Nice work (almost). Your approach seems like a reasonable approximation, for when the base of your object, traveling along the ground, strikes a hard stop (an obstruction) on the ground, provided we neglect strain energy temporarily stored at the contact site (which I did not want to include, because we would need the mass distribution of the striking body, and the stiffnesses of the two colliding entities at the contact site, not to mention it would be too detailed). You, however, made one minor mistake; you forgot to multiply by gravitational acceleration, g. Therefore, your current energy balance equation should instead be 0.5*m1*v1^2 = m1*g*h, where v1 = impact velocity, h = max[0, (6.09^2 + 0.79^2)^0.5 - (6.09/cos(theta))] m, and theta = downward slope angle of ground surface. Solving the above energy balance equation for v1 gives, v1 = (2*g*h)^0.5. According to this analysis, if the velocity of your object equals or exceeds v1, your object will tip over when it strikes the obstruction.

    Therefore, if the ground surface is horizontal, theta = 0 deg, h = 0.051 026 m, and v1 = [2(9.81 m/s^2)(0.051 026 m)]^0.5 = 1.001 m/s. If theta = 5 deg, then h = 0.027 763 m, and v1 = 0.738 m/s. If theta = 7.4 deg, then h = 0 m, and v1 = 0 m/s. In other words, if the ground surface slopes downward 7.4 deg, your object tips over even when its velocity is zero.

    The tipping force required to tip the object over would be F = m1*g*tan(7.39 deg - theta), where m1 = object mass, and theta was defined earlier. Note that if F is negative, no additional force, other than gravity, is required to tip the object over.
     
  4. Oct 14, 2009 #3
    Hi NVN,

    that is a great response, I understand where you are coming from, i was going to ask before your second post how i could calculate the force required to tip at an angle, however you have already answered this for me.

    Thankyou kindly, I really appreciate this!

    I can now move onto the next problem it seems.

    Cheers

    Glocki35
     
  5. Oct 25, 2009 #4

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    glocki35: To answer your PM question, the 7.39 deg angle in post 2 comes from the given object geometry, shown in your diagram. The angle will be different for different objects. The angle depends on the object geometry. For different objects, determine this angle in a manner similar to the method you used in your post 1 diagram.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Potential Energy & Tippipng Forces
Loading...