What forces are required to tumble a block?

[abrogard]

TL;DR Summary

want to know the effort required to move blocks by hand 'tumbling' them over and over

I just want to know how strong I have to be to move a large stone block by tumbling it over.
A formula is what I'm looking for I guess.
Assume a perfect cube.
Begin with a raised edge making say 10 degree angle with the ground.
I get my fingers in there and start lifting.
I suppose when I get to 45% the thing will be balanced and the tiniest force will tip it over.
I also suppose at in the beginning there at 10 degrees I'll need to lift almost the total weight of the thing.
We've got sandstone blocks here in all shapes and sizes got us thinking about it.
Not easy for us to figure: we're not good on maths.
Can't even be sure the change in needed power would be straight line between 'nearly all' and 'nearly none'.
I'm thinking it should be a straight line and I'm thinking it will be the same percentage of the weight of the block for all perfectly cubic blocks with homogeneous composition.
And I'm thinking it will turn out to be a very simple equation.
And I'm thinking long rectangular blocks are essentially a lot of cubes stuck together so the formula could obviously be applied to them. Just multiply by the number of cubes you can fit in.
That's all we know.

 

[Baluncore]

I also suppose at in the beginning there at 10 degrees I'll need to lift almost the total weight of the thing.

You only need to lift half the weight, since the opposite edge rests on the ground, as a fulcrum. You also have a 2:1 lever between the fulcrum, the centre of mass, and your hands. The first bit is the hardest.

It is easier to move heavy blocks of rock underwater.

 

[Halc]

I also suppose at in the beginning there at 10 degrees I'll need to lift almost the total weight of the thing.

At 0 degrees you'll be lifting half the weight, the other half being born by the far edge.
At 10 degrees, a substantial majority of the weight lays on the far edge.

Can't even be sure the change in needed power would be straight line between 'nearly all' and 'nearly none'.

Power depends on how fast you do it. It theoretically can be done with negligible power no matter the size of the block, if you're willing to do it slowly.
But if you do it quicker, there is inertia to consider, and the continued force needed to tip it might go to zero well before the center of mass passes over the far edge.

And I'm thinking it will turn out to be a very simple equation.

For vertical edge lifting, it's just a matter of computing where the center of mass is relative to the two edges. If it is centered, half the weight is on each edge. At 45 degrees, the CoM is 0% of the distance between the far edge to the edge being lifted, so zero weight where you lift.
It all forms a triangle, so the answer is trigonometric, a function of width and height.
It depends on the shape of the block. A tall block might already tip at 10 degrees. A really long low block will need almost 90 degrees before the near edge bears no weight.

(Question to others) This is not in the homework section. Am I allowed to give more detail?

 

[abrogard]

the block I'm hypothesizing is a perfect cube. our blocks on the job are not. this is not homework. I'm not a child, far from it. just a rather mathematically illiterate adult.
Re: your question to others - I would like sufficient detail to (1) give me a formula that one applies to this question and (2) enable me to understand it rather than merely use it.
re 'tall block' etc - I say again: postulated is perfect cube.

 

[Filip Larsen]

The mechanical work required to rotate the block to 45 deg can be approximated by $E = m g h$, where $m$ is the mass of the block, $g$ is the gravitational acceleration (9.8 m/s²) and $h$ is the height the center of mass is lifted.

For a cubic block the change in height of the center of mass is $h = s / \sqrt{2}$ at 45 deg with $s$ being the side length of the cube and mass can now be written as $m = \delta s^3$, with $\delta$ being the density, i.e.
$$E = \frac{\delta g}{\sqrt{2}} s^4.$$
By the way, if you can lift a maximum of $m_*$ straight up, then this relates to the maximum block size $s_*$ you can begin to tilt as $m_* = \tfrac{1}{2}\delta s_*^3$, that is, $$s_* = \sqrt[3]{\frac{2m_*}{\delta}}.$$

For example, for sandstone using a $\delta$ of 2500 kg/m³ and $m_*$ of 100 kg, this would give an $s_*$ of 43 cm and cost you at least around 600 J (provided I punched in the numbers correctly, haven't had my morning coffee yet).

 

[Lnewqban]

the block I'm hypothesizing is a perfect cube. our blocks on the job are not. this is not homework. I'm not a child, far from it. just a rather mathematically illiterate adult.
Re: your question to others - I would like sufficient detail to (1) give me a formula that one applies to this question and (2) enable me to understand it rather than merely use it.
re 'tall block' etc - I say again: postulated is perfect cube.

How heavy is that actual cube?
What material it is made of?

What means do you have to lift and turn the block over its edge?
Can you just let it drop naturally on the opposite side without causing damage to its face or surface?
If not, could you use the same lifting mechanism to slow it down in the way down?

 

[jrmichler]

TL;DR Summary: want to know the effort required to move blocks by hand 'tumbling' them over and over

I just want to know how strong I have to be to move a large stone block by tumbling it over.
A formula is what I'm looking for I guess.
Assume a perfect cube.
Begin with a raised edge making say 10 degree angle with the ground.
I get my fingers in there and start lifting.
I suppose when I get to 45% the thing will be balanced and the tiniest force will tip it over.
I also suppose at in the beginning there at 10 degrees I'll need to lift almost the total weight of the thing.
We've got sandstone blocks here in all shapes and sizes got us thinking about it.

Not so much a formula, as a calculation procedure. Assume the crudely sketched block in the figure below is on level ground, has a weight $W$, and that you apply the tipping force as $F_1$, $F_2$, $F_3$, or $F_4$. We solve these problems by taking a sum of moments. A moment is a force multiplied by a distance. The weight of the block $W$ is a force straight down, and that force is through the center of the block. You can apply your tipping force by pushing horizontally at a corner, pushing vertically at a corner, pushing at an angle at a corner, or pulling on a corner.

Assume the (perfect cube) block has length / width / height $L$, and the tipping point is the lower right corner. We take our moments about the tipping point. The moment due to the weight $W$ is the weight times the perpendicular distance from that force to the tipping point. That distance is $L/2$, so the moment is $W * L / 2$, or $WL/2$. Similarly, the moments due to tipping forces $F_1$, $F_2$, or $F_4$ is $F_1L$, $F_2L$, or $F_4L$. When the tipping moment is equal to the moment due to the weight of the block, it tips. The formula is $F_1L = WL/2$, which we solve for the force $F$ and get $F = W/2$.

The force $F_3$ is at a 45 degree angle, so the perpendicular distance is $1.414 * L$, and the tipping force is $F_3 = 0.35L$. And you are correct, when the block is at 45 degrees, it will be balanced. Sandstone weighs 100 to 170 lbs per cubic foot depending on porosity and water content if you want to estimate the weight of your blocks.

 

[Baluncore]

Just remember the Health and Safety, Proverb 26:27
"Whoever digs a pit will fall into it,
And he who rolls a stone will have it roll back on him."

 

[abrogard]

as

Not so much a formula, as a calculation procedure. Assume the crudely sketched block in the figure below is on level ground, has a weight $W$, and that you apply the tipping force as $F_1$, $F_2$, $F_3$, or $F_4$. We solve these problems by taking a sum of moments. A moment is a force multiplied by a distance. The weight of the block $W$ is a force straight down, and that force is through the center of the block. You can apply your tipping force by pushing horizontally at a corner, pushing vertically at a corner, pushing at an angle at a corner, or pulling on a corner.

Assume the (perfect cube) block has length / width / height $L$, and the tipping point is the lower right corner. We take our moments about the tipping point. The moment due to the weight $W$ is the weight times the perpendicular distance from that force to the tipping point. That distance is $L/2$, so the moment is $W * L / 2$, or $WL/2$. Similarly, the moments due to tipping forces $F_1$, $F_2$, or $F_4$ is $F_1L$, $F_2L$, or $F_4L$. When the tipping moment is equal to the moment due to the weight of the block, it tips. The formula is $F_1L = WL/2$, which we solve for the force $F$ and get $F = W/2$.

View attachment 337430

The force $F_3$ is at a 45 degree angle, so the perpendicular distance is $1.414 * L$, and the tipping force is $F_3 = 0.35L$. And you are correct, when the block is at 45 degrees, it will be balanced. Sandstone weighs 100 to 170 lbs per cubic foot depending on porosity and water content if you want to estimate the weight of your blocks.

as best I can make out you're about on the same page as me.
I seem to have failed to spell out some things: made assumptions.
I want everything at its simplest.
remember i'm simply wondering how much muscle we men would have to put in to tumble blocks over.
a very 'low level' ordinary man thing. not some fancy high level theoretical physics thing.
like we can assume that forces will always be applied in the 'best' direction or if that's to complicating of things then just have them applied along the face in question. See what I mean? Like in the beginning we lift straight up and I'm calling that 'across' or 'along' the face we are looking at. And we can make that same assumption, can't we, for all future applications of the force.
We don't care about the density of sandstone or anything else as I understand it.
Because the way I see it, instinctively, the force required will always be proportionate to the weight and should be capable of being expressed that way. So we get a general rule applicable to all homogeneous cubes.
Is that right?
Can that be done?
I might be mis-stating or misunderstanding the whole thing. I don't know. I'm thinking just as like when a man's working. You get your fingers underneath and start lifting and you keep applying force until it tumbles over.
Now what's that? Application of the same force throughout the whole process? Doesn't feel like it. Seems difficult at first and then easier and easier until suddenly it falls over.
Seems we just need a force at any time that is sufficient to move it.
And that force varies as the angle of tilt increases.
I'm looking for a 'rule' or formula that I could, for instance, put into a python script and it would output to me the forces required at any particular angle. And I'm wondering if I graphed them would it be a straight line from a max (which as I say would be directly related to the weight of the block I suppose) to the min at the 45 degree balance point - or would it be a curve?
I wonder that. And I wonder what's the 'formula'.
p.s. I notice your sketch seems to have the first force applied at the top at 45 degrees. This is not what I was/am thinking of. I imagine a first force as applied upwards from the bottom corner. That's why I started by saying let there be a 10 degree angle so's you can get your fingers in there. Not needed, I know, for mathematical problems etc - but just making it 'real'.

 

[jbriggs444]

If I were trying to roll a cube of decent size with maximum effect, I'd get finger tips below to anchor and also push with my chest. There would be two forces applied. Or one net force plus a net torque if you wanted to decompose things that way.

 

[A.T.]

...we can assume that forces will always be applied in the 'best' direction...the forces required at any particular angle. ....

For cubes of uniform density and force applied in optimal direction at the bottom edge opposite to pivot edge (or the edge above the pivot edge):

F = m*g*cos(alpha + pi/4) / sqrt(2)

Force applied in optimal direction at the edge diagonally opposite to pivot edge :

F = m*g*cos(alpha + pi/4) / 2

where alpha is the tilt angle in radians.

 

[abrogard]

Thank you for that. I think with a little help from chatgpt I can get a procedure out of that. I think we can work on either of the two options for where the force is applied. It will be interesting to see what differences.
In practice we find in the physical environment we generally mainly exert the force throughout from the one place. Like if we get fingers under an edge we then lift that edge and keep applying from there until the block turns over.
I never specified but of course I mean just one 'turn' and that actually moves the cube through 90 degrees only.
If possible - like two men on the job or something - it may be that one hand will push at the 'upper edge' while the other hand is lifting the lower edge. So there we have both in operation.
I'm hopeful about these formula. Thanks.

 

[Baluncore]

If possible - like two men on the job or something - it may be that one hand will push at the 'upper edge' while the other hand is lifting the lower edge. So there we have both in operation.

Two people, employing brute force and fingers, will sometime lack coordination. It is inevitable that complacency will set in. When that happens, one will jump out of the way, while the other will be hurt, in a foreseeable industrial accident.

It is safer to lift one edge of a rock, with a pry bar, using a small wooden block as a fulcrum. Then a longer bar can slide under the rock, for use as a lever to turn the block. You can slide blocks or wedges under the rock as you open the gap for the longer lever. That way, you never need to place any part of your body under the load.

 

[Baluncore]

In practice we find in the physical environment we generally mainly exert the force throughout from the one place. Like if we get fingers under an edge we then lift that edge and keep applying from there until the block turns over.

No, we do not.
That is too simplistic a physical model to explain the technique used by a skilled and experienced operator.

Lifting with your fingers alone, will require you to place more of your body, under the load. Your fingers will slip as the block starts to turn, and the finger-bend-angle reduces from 90° toward 45°. It all comes down to the tangent of the contact angle, becoming equal to the coefficient of friction. Slippery fingers will not slide if you can keep them wrapped at 90° around the lower edge of the rock.

Instead, you should hook your fingers under the lower edge, then lift to increase the down force on your feet, for improved traction on the ground. Then by pushing with your upper-arms or shoulders on the top edge, you can then turn the rock to 45°. That keeps your fingers bent at 90° around the lower edge, so they do not slip. In effect, your upper-body rolls with the near-face of the block, with your legs doing the work, not your back. As the rock reaches the balance point near 45°, you push off and step well back, before the rock has a chance to roll back on you, in accordance with proverb 26:27.

 

[abrogard]

For cubes of uniform density and force applied in optimal direction at the bottom edge opposite to pivot edge (or the edge above the pivot edge):

F = m*g*cos(alpha + pi/4) / sqrt(2)

Force applied in optimal direction at the edge diagonally opposite to pivot edge :

F = m*g*cos(alpha + pi/4) / 2

where alpha is the tilt angle in radians.

Thanks again for that. I got a routine up and finished up with a line graph of the results. I was surprised to find the top edge so far in front.

 

[A.T.]

I was surprised to find the top edge so far in front.

The diagonal of a square is sqrt(2) times longer than its side, so the force with diagonal lever arm is 1/sqrt(2) times the force with bottom lever arm.

 

[abrogard]

I see. Thanks. I perhaps faintly begin to understand something about this basic aspect of physics. :)