Potential Energy & Tippipng Forces

In summary: That is, find the intersection of the line segment that passes through the center of mass (CM) of the object and the ground, and then use the cosine of this angle to calculate the tipping force.
  • #1
glocki35
11
0
Hi All,

I have a problem i am trying to figure out, if an object (scale drawing in attachment) is traveling 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.
 

Attachments

  • Scale Drawing.pdf
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  • #2
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.
 
  • #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
 
  • #4
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.
 

1. What is potential energy?

Potential energy is the stored energy an object has due to its position or state. It is the energy that an object has the potential to convert into other forms of energy, such as kinetic energy.

2. How is potential energy related to tipping forces?

Potential energy is related to tipping forces because an object's potential energy increases as it is lifted or tilted, creating a tipping force that can cause the object to fall or topple over.

3. What are some examples of potential energy and tipping forces in everyday life?

Some examples include a book sitting on a shelf (potential energy) and a person tipping over a glass of water (tipping force).

4. How is potential energy and tipping forces calculated?

Potential energy is calculated by multiplying the object's mass, gravity, and height. Tipping forces can be calculated by considering the center of mass of an object and the direction of the force being applied.

5. How does potential energy and tipping forces play a role in engineering and design?

Potential energy and tipping forces are important considerations in engineering and design because they can affect the stability and safety of structures and objects. Engineers and designers must take into account potential energy and tipping forces to ensure that their creations are structurally sound and do not pose a risk to people or the environment.

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