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Toppling force calculation

by BertSmurf
Tags: calculation, force, toppling
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BertSmurf
#1
May2-12, 12:57 AM
P: 8
I am attempting to design a base to support a post that will be pulled horizontally at its top.
A 2D image of the post will be a simple inverted "T"
The top of the post will be pulled horizontally by a force of approximatly 900N
What I need to do is design a suitable base for this post that will prevent it from toppling (suitable weight and dimension)
I am a little confused on the mathematics that govern the force required to topple something like this.
A moment balance about the point around which the post will topple (the edge of the base) seems to imply that the problem is independent of the COG (since the prependicular distance wont change as COG is in the middle) and this cant be right.
Can some one please shed some light on this?
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tiny-tim
#2
May2-12, 05:03 AM
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Hi BertSmurf! Welcome to PF!
Quote Quote by BertSmurf View Post
A moment balance about the point around which the post will topple (the edge of the base) seems to imply that the problem is independent of the COG (since the prependicular distance wont change as COG is in the middle) and this cant be right.
No, it is right

for a fixed mass, the height of the centre of mass will not affect the force needed to start it toppling.

(but obviously, once it moves, the higher the centre of mass, the easier it becomes )
PhanthomJay
#3
May2-12, 04:27 PM
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The line of action of the posts weight thru its com is perpendicular to the pivot at the edge of the base. The wider the base, the harder it is to topple.

BertSmurf
#4
May2-12, 06:00 PM
P: 8
Toppling force calculation

Thanks for the help guys. Now that I look at it, it seems obvious enough! :)
BertSmurf
#5
May3-12, 03:10 AM
P: 8
Ok, I seem to have stumbled upon a bit of a contradiction here.
I took moments the centre of the base as the post was just about to tip over (so the line of action was directly above the point over which the whole post assembley would rotate).
The forces acting on the post are the weight of the post (W) acting down at the centre of gravity (a distance HCOG from the base) and a horizontal force (F) is acting at the top of the post (a height Hpost from the base).
The post is at an angle λ to the ground. (my assumed x and y planes are at this same inclination for simlicity of calculations)
The inclined x component of the Weight of the Post is Wsinλ*HCOG
The inclined x component of the applied force is Fcosλ*Hpost
Summing these moments gives the equation Wsinλ*HCOG+Fcosλ*Hpost=0
Resolving for F gives: F=-Wsinλ*HCOG/Fcosλ*Hpost
Looking at this equation means that as the COG is higher, the required force increases and vice versa.
This is surely wrong....
Can someone please point me in the right direction, I am making a wrong assumtion somewhere but cant figure it out.
tiny-tim
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May3-12, 06:40 AM
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Hi BertSmurf!
Quote Quote by BertSmurf View Post
I took moments the centre of the base as the post was just about to tip over (so the line of action was directly above the point over which the whole post assembley would rotate).
But won't the applied force F then be 0 ?
PhanthomJay
#7
May3-12, 02:42 PM
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I think you are losing sight of the question with all this alpha stuff and trig.

I believe all you are being asked is to determine a suitable weight and dimension of the base plate such that the moment about one edge of the plate from the horizontally applied force at the top is equal to the counteracting moment of the base's weight about that point. You must be given the height of the post to do this. The weight of the post is not given, so unless it is, you might just neglect it and assume all the weight is due to the base plate and acts vertically down at the center of the post/plate. You have more than one solution to the problem, since you can use a very heavy plate with a small width extending from the center of the post, or a lighter plate with a large width extending from the center of the post.
For example, if the applied force at the top was 1000N and the post was 1 meter tall, and the post weight was negligible, then the overturning moment about one edge (the pivot point at the instant where tipping starts) would be 1000 N-m, so you would have to adjust the base plate Weight (W) and the horizontal distance from the plate centerline to the edge of the plate, X, such that (W)(X) = 1000 N-m. Unless I've misunderstood.
BertSmurf
#8
May3-12, 05:12 PM
P: 8
You have made that very clear thanks Jay.
Im actually kicking myself right now... :)


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