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

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Discussion Overview

The discussion revolves around calculating the force required to topple a frame with a conveyor attached, focusing on the mechanics involved, including the center of gravity, angles, and the role of swivel castors. Participants explore various methods and considerations related to this problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the mass of the unit as 175 kg and calculates the vertical force as 1715 Newtons, suggesting methods involving torque and angles for determining the force needed to topple the frame.
  • Another participant calculates the force required to push on the 16° side and the 12.5° side, providing specific values of 380.2 N and 491.8 N, respectively, based on the tangent of the angles.
  • Further calculations are provided regarding the height of the center of mass and the corresponding torque values for the forces applied, with formulas for force based on height.
  • A participant inquires about how to factor in the swivel castors into the calculations, seeking a simpler method for consideration.
  • Another participant suggests that the swivel castors could be affected by small objects on the floor, which may block their movement, indicating a potential complication in the calculations.

Areas of Agreement / Disagreement

Participants present various calculations and methods, but there is no consensus on how to incorporate the swivel castors into the analysis or on the best approach to calculate the required force. Multiple competing views and methods remain in the discussion.

Contextual Notes

Participants express uncertainty about the impact of swivel castors and the conditions under which they may not function as expected, highlighting potential limitations in their calculations.

D19A99G
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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,
 

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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.
 
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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.
 

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