Find Angle Needed for Objects to Topple Over - Tom

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To determine the angle at which the larger box will topple over, the torque created by the weight of both boxes must be calculated. The center of gravity for each box plays a crucial role in this calculation, as the torque is the product of the weight and the horizontal distance from the center of mass to the pivot point. If the center of gravity of the top box extends beyond the pivot point, it will topple; otherwise, it will return to its original position. The critical angle occurs when the diagonal of the top box becomes vertical. The discussion emphasizes that understanding the physical setup and performing basic calculations are essential for solving the problem.
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If I have a box 1m deep by 2m high weighing 50kg and then bold another box onto it at one end that is 0.275m deep by 2.5m high weighing 100kg.

Assuming each objects centre of gravity is central. At what angle would I need to lift the larger box to so pivoting on the smaller boxes corner before it would topple over with the help of gravity and not just fall back to its starting position?
Thanks
Tom
 
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Welcome to PF!

Hi Tom! Welcome to PF! :smile:

Use the rotational version of good ol' Newton's second law … τtotal = Iα

for α to be > 0, τtotal must be > 0,

(you needn't worry how much I is, it doesn't make any difference to whether Iα > 0, does it? :biggrin:)

so calculate τ (the torque or moment) for the weights (mg) of each of the two centres of mass :wink:
 
Sorry to sound stupid but how would i calculate this out then as i am not that brilliant at maths
tom
 
tomloaf said:
Sorry to sound stupid but how would i calculate this out then as i am not that brilliant at maths
tom

Good old wikipedia can help. Follow these short steps to enlightenment!

Type "www.wikipedia.org" in your browser.
Type the following "Mass Moments of Inertia"
Read article.

The subsequent article shall detail all prerequisite knowledge and information needed to calculate the center of gravity of your system.

Just a footnote: Extensive mathematics is not required for such an operation, short of addition, multiplication and/or division. Only knowledge of the physical system, i.e. physics.

Have a great day!
 
hi tom! :smile:
tiny-tim said:
so calculate τ (the torque or moment) for the weights (mg) of each of the two centres of mass :wink:
tomloaf said:
Sorry to sound stupid but how would i calculate this out then as i am not that brilliant at maths

the weight acts vertically through the centre of mass …

so the torque (the moment) is the weight times the horizontal distance from the centre of mass to the pivot point (ie the edge it's about to tip over on) :wink:
 
If you are only tilting the top box, like it seemed to me from the description, then you only need to care for its center of gravity and the parameters of the bottom box are completely irrelevant to the problem. Which is weird, so I am assuming I didn't really understand the setup.
But still, here's my stab at it: I don't think you need to calculate torque or **** at all, just project the center of gravity of the top box straight down. If it is inside the base then the box will fall back to its base and be stable, if it is outside it will topple over. So the critical angle is the one when the box's diagonal section becomes vertical.
 
The set up is like this
I need to know the angle the indicated point can be raised before it topples over
Tom
 

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tomloaf said:
I need to know the angle the indicated point can be raised before it topples over.
16.70 deg
 
sorry to sound like a teacher but can you show your working as i have some other calculations like this
tom
 
  • #10
ok find the combined center of gravity, then see what the angle is when it is right above the pivoting edge.
 
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