Force required to tip over an item 6

[jonesy01]

Need help, due to safety issues I need to find out what force would be needed to tip over a cart.

I am analyzing force required to tip over a cart and the cart consists of 3 items:
-The cart with wheels = (148 lbs)
-Tank w/water = (370 lbs)
-Pump = (20 lbs)

The way I am analyzing this is if the front wheel collides with something what’s the force required for the cart to tilt or tip forward. I believe the cart will tip if the torque at the wheel exceeds the torque due to the total weight of the cart.

-1st, I found the center of gravity for the cart with all components. Multiplied the weight of each component times their midpoint distance from the front of the cart to get their moments. Added them all together.

pump, 3.87 in x 20 lb = 77.4 lb/in
tank, 20 in x 370 lb = 7400 lb/in
cart, 18.5 in x 148 lb = 2738 lb/in
----------------
10215.4 lb/in

-2nd, got the sum for the weight of the cart and all components = 538 lb and then divided the moment sum by the weight and got 18.98 in as the center of gravity from the front of the cart.


-3rd, the only way the cart will tip is if the torque at the front wheels exceeds the torque due to the weight of the cart. Formula I used:

TORQUEcw = TORQUEccw

(f)(d) = (f)(d)

(f)(41.56”) = (538 lb)(18.98”)

f = 245 lb, so 245 lbs needed to tip over the cart if the front wheels at point A hit something and if the cart with all items weighs 538 lbs.

Am I missing anything here. Any advise is much appreciated.

 

[Strong]

A quick look at your moment table indicates that the center of gravity of each component does not appear to be referenced to the same location/elevation, ie. the floor. I assume you are looking for the composite CG of the three items above the floor??

Walt

 

[drawoh]

AVxENG,

I asked this question in the Structural Engineering forum.

thread507-255357

JHG

 

[Denial]

A few comments:

(1) The general approach is simplistic, but might still be appropriate in some circumstances.

(2) A structural engineer would apply a couple of safety factors into your formulae, one to increase the forces tending to cause overturning, and the other to decrease the forces tending to resist overturning.

(3) Where did the "d" value of 41.56" come from?

(4) Expanding on the previous point. What is the source of your anticipated overturning force? Is it because the donkey keeps pulling at the cart? Is it an inertia force from a rapid deceleration of the cart when it hits the obstacle?

(5) If it is an inertia force, you need an estimate of the deceleration. This in turn will depend upon how fast the donkey was pulling at the time of impact, how flexible the various parts are, etc. The amount of info needed and the amount of computation involved can rapidly get out of control.

(6) One item that will definitely be "flexible" is the water in the tank. Not a problem if the tank is completely full (or completely empty), but it is easy to envisage a situation where a partially full tank is the worst case. For the simple "statics" approach you have used, a conservative way to allow for the flexibility of the water might be to assume that the water is all at the front of the tank rather than in the bottom of the tank (with the water surface vertical rather than horizontal).

(7) A totally alternative approach to the entire problem would be to consider energies rather than forces. Before impact you have only kinetic energy (KE). After impact (assuming everything — except the water — stays rigid) the entire system begins to rotate about the front axle. As it rotates, its speed of rotation slows down because your KE is being converted into (gravitational) potential energy (PE). Eventually it stops rotating, when all the KE has become PE. It should be easy to calculate the rotation angle reached under this set of assumptions. Is this angle beyond the angle at which the cart begins to summersault? This simplified energy approach is VERY conservative, because in practice a large proportion of the KE will be converted into other forms of energy (heat, noise, water sloshing, elastic strain energy, inelastic strain energy,…). However it might help you to get a rough feeling for things.

(8) I have not even started to check your arithmetic.

HTH

 

[msquared48]

For a proper design, the resisting moment must be a minimum of 1.5 times the overturning moment.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[desertfox]

Hi AVxENG

If your cart is only moving very slowly then a static calculation on tipping should give a reasonable answer to your question, however if its moving at speed its a different story because you have acceleration involved and possibly sloshing of water in the tank.
I'll just point out that your torque figures should say lbs in and not lb/in, also in your analysis you need to take moments about the axle of the wheel thats just hit a dead stop and not the front of the cart.
So whilst I agree how you found the horizontal position of the resultant CofG I believe you should take the 18.98" dimension off the distance from the front of the cart to the centre of the front wheel for your lever arm,the force to tip the cart should be the vertical distance from the the centre of the front wheel and not the floor, see my attached sketch.

desertfox

 

[jonesy01]

Strong: The answer to your question is yes.

 

[jonesy01]

Denial: The dimension of 41.56" comes from the design of the cart. This is a dim we have assigned. As for your next question, The force is derived from someone pushing the cart forward and hitting an object. The person's continued forward force would overturn the cart. It is not due to a deceleration. Also the tank is full

 

[jonesy01]

Desertfox: so from your sketch my formula should be:

F x 41.56 = 538 x (distance from CoG to midpoint of wheel)

Solve for "F"

Thanks for the sketch but I'm not fully understanding your logic.

 

[desertfox]

hi AVxENG

If someone is pushing the cart from right to left, then suddenly the front wheel locks up the trolley will try to rotate in a clockwise direction and the pivot point will become the front wheel axle, the only thing stopping the cart from turning over is the total weight acting down from the pivot point, if the person can overcome this resisting turning moment the trolley starts to tip.
Look at where your taking your moment from ask yourself why would the trolley tip about about the back edge of the trolley where the person is pushing it, if you push a supermarket trolley and the front wheels jam while your pushing what tends to happen and in which direction does the trolley try to tip.

desertfox

 

[rb1957]

you're looking at a case where the cart hits a step and stops ... where is the reaction force from the step applied to the cart ? (not at ground level). your diagram is a good start towards a free body diagram, complete it to understand the situation.

other good points above ... safety factors, partially full tank (a simple graph will show the change in inertial moment and CG to resolve this question).

 

[jonesy01]

desertfox: Thanks for the response. Sorry I was not clear. I understand what you meant by saying the pivot point would be the front wheel axle I just did not fully understand your sketch. According to your sketch I'm thinking my formula will be:

F * 41.56 in = 538 lb * (distance from CoG to midpoint of wheel)

Solve for "F"

Your input is much appreciated. Thanks again.

avxeng

 

[zekeman]

You do this problem with energy exchange. All other methods are unnecessarily complicated and involve too many assumtions.

First, locate the center of mass of the system. Next, get a pencil and draw a line between the CM and the front axel.

Now just before impact of the front wheel, the total energy of the system referred to the centerline of the front wheel is

1/2MV1^2+Mg*h1

where
M= total mass
V1 initial velocity
Mgh1= initial potential energy


Now ,assuming the energy is conserved (in this case a very conservative assumption), you next equate this to the energy of the system after the collision rotating the wagon to a new CM height, h2

1/2MV1^2+MGH1=1/2MV2^2+Mgh2

Now in order for the cart to topple,h2 must achieve the
the highest elevation possible . Further the minimum amount of energy for toppling would coincide with V2=0
Now the highest point above ground, h2 is the distance you drew from the CM and the front axle centerline. So, the minimum KE for toppling would be

1/2mMV^2=Mg(h2-h1)

 

[desertfox]

Hi AVxENG

Your last post is correct except that your 41.56" diatance is wrong, tou need to take the 41.56" and minus the vertical distance to the front wheel axis.

To use the energy method and do it properly you would need to note you need to calculate the angular kinetic energy of the rotating wheels as well as the kinetic energy of the cart's forward motion unless you ASSUME its negligible, then you need to know the velocity of the cart and the mass of the person pushing it as he will be involved in the collision, then finally you ASSUME
there are no energy losses to enable yourself to find the height at which your centre of gravity as moved vertically and hence obtained a force.
I think your initial approach AVxENG using a static approach for a slow moving cart is the best and certainly easier to calculate.

desertfox


desertfox

 

[rb1957]

if you drew a free body, you'd've seen desertfox's point ...