How Much Force Is Needed to Topple a Frame on Castors?

  • Thread starter Thread starter D19A99G
  • Start date Start date
  • Tags Tags
    Frame Toppling
AI Thread Summary
To determine the force needed to topple a 175 kg frame with a conveyor, calculations show that pushing on the 16° side requires approximately 380.2 N, while pushing on the 12.5° side requires about 491.8 N. The torque generated by these forces is 377.2 N·m and 487.9 N·m, respectively, when factoring in the height of the center of mass. The discussion also highlights the importance of considering swivel castors, which may become obstructed, affecting stability. Understanding these dynamics is crucial for ensuring safety and preventing industrial accidents.
D19A99G
Messages
4
Reaction score
1
Hello all,
I'm trying to calculate how much force is required to topple over a frame with a conveyor attached to it. I have attached a screenshot of the assembly.
I have the centre of gravity as seen on the screenshot. The two lines to the pivot points show the angle. These are 16 degrees (line length of 1032mm) and 12.5 degrees (line length of 1015mm).
The total mass of the unit is 175kg.

I have seen a couple of methods online such as using kinetic and potential energy or trying to calculate the torque required to push the CoG over the pivot.

I'm struggling with this seemingly simple task! Any input would be appreciated, if anymore information is required let me know.

Thanks,
 

Attachments

  • Screenshot.PNG
    Screenshot.PNG
    20.1 KB · Views: 188
Engineering news on Phys.org
Welcome to PF.
Mass is 175 kg. Vertical force = m·g = 175 * 9.8 = 1715 Newton.
Push on the 16° side, fall towards 12.5° side. 1715 * Tan( 12.5°) = 380.2 N
Push on the 12.5° side, fall towards 16° side. 1715 * Tan( 16.0°) = 491.8 N

Edit: The force is assumed to be applied to the centre of mass.
 
Last edited:
  • Like
Likes D19A99G and dlgoff
Height of the centre of mass is; 1.015 m * Cos( 12.5° ) = 0.992 m
Check height of centre of mass; 1.032 m * Cos( 16.0° ) = 0.992 m

Push on the 16° side, fall towards 12.5° side. 1715 * Tan( 12.5°) = 380.2 N.
Which represents a torque of 380.2 N * 0.992 m = 377.2 N·m.
If the force is applied at height of h metre, the force in Newtons will be; F = 377.2 / h Newton.

Push on the 12.5° side, fall towards 16° side. 1715 * Tan( 16.0°) = 491.8 N.
Which represents a torque of; 491.8 N * 0.992 m = 487.9 N·m.
If the force is applied at height of h metre, the force will be; F = 487.9 / h Newton.
 
Thank you for the response!
How would the swivel castors then be factored into that? Is there any easy way?
 
One failure is one industrial accident too many. The easiest way to factor in the swivel castors is to assume the castor will sometimes be blocked from turning or rolling by a small object on the floor. That happens more often than you might expect.
 
Here's a video by “driving 4 answers” who seems to me to be well versed on the details of Internal Combustion engines. The video does cover something that's a bit shrouded in 'conspiracy theory', and he touches on that, but of course for phys.org, I'm only interested in the actual science involved. He analyzes the claim of achieving 100 mpg with a 427 cubic inch V8 1970 Ford Galaxy in 1977. Only the fuel supply system was modified. I was surprised that he feels the claim could have been...
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
Back
Top