Calculating max weight object supports without counterweigh

[James O'Neill]

In tower cranes counterweights are placed on the counter jib to prevent the crane from falling over. Whereas on mobile cranes the counterweights are placed directly behind the cabin.
The crane in itself also acts as some counterweight.
My question to you is:
How much weight can an object support on one side, without using counterweights, before falling over?
E.G. a 40 cm arm with an 8 Kg weight is added to a 20 Kg box. Will the box support the weight without tipping to the weighted side?

 

[BvU]

Depends on the size of the box. You want to make a balance about the potential tipping axis.
On the left you have 20 kg x half the length of the box
On the right you have 8 kg x 0.4 m

Half the length applies if the box has uniform mass density (horizontal distance between center of mass and tipping axis -- the lower right)

Make a force diagram to see.

 

[James O'Neill]

I have made a sketch as example.

 

[BvU]

So on the lefet you have 20 x 0.125 kg m and on the right 8 x 0.4 kg m. Sum is 0.7 kg m to the right will keel it over !

 

[James O'Neill]

I'm having some trouble seeing how you got 0.7 kg m at the end??
Also just curious, does three dimensional properties (l x h x w) have any effect on this?

 

[BvU]

I'm having some trouble seeing how you got 0.7 kg m at the end??
Also just curious, does three dimensional properties (l x h x w) have any effect on this?

20 x 0.125 kg m minus 8 x 0.4 kg m is minus 0.7 kg m.

Also just curious, does three dimensional properties (l x h x w) have any effect on this?

Of course. If you lay the box on its side, you get 20 x 0.25 kg m minus 8 x 0.4 kg m is plus 1.8 kg m and the contraption will stay put.

 

[James O'Neill]

20 x 0.125 kg m minus 8 x 0.4 kg m is minus 0.7 kg m

Ohhh... I thought since you said sum in your second message I had to add the two together.

If you lay the box on its side, you get 20 x 0.25 kg m minus 8 x 0.4 kg m is plus 1.8 kg m and the contraption will stay put

This makes sense.
If the 8kg weight was directly against the side of the box, it would also stay up, right? (20 x 0.125 kg m) - (8 x 0.125 kg m) = 1.5 kg m.

 

[BvU]

Ohhh... I thought since you said sum in your second message I had to add the two together.

Yes, sum They are vectors and try to rotate in opposite directions, so one has a minus sign. Analogous to pulling to the left and pulling to the right.

If the 8kg weight was directly against the side of the box, it would also stay up, right? (20 x 0.125 kg m) - (8 x 0.125 kg m) = 1.5 kg m.

Nice try, but that case the 8 kg would try to rotate to the right with 8 kg times the horizontal distance of the center of mass of the 8 kg weight to the potential tipping point. Which is not the .125 m if it's an iron weight

 

[James O'Neill]

So let's make the weight 10 cm long.
(20 x 0.05 kg m) - (8 x 0.05 kg m) = 0.6 kg m. Is it right this time?
Can you then adjust this formula to calculate counterweight needed?

 

[BvU]

So let's make the weight 10 cm long. (20 x 0.125 kg m) - (8 x 0.1 kg m) = 1.7 kg m. Is it right this time?

It would be (20 x 0.125 kg m) - (8 x 0.05 kg m)

See it as a see-saw or a weighing scale with uneven arms. As long as m₁x₁ - m₂x₂ > 0 it won't topple over point P

 

[James O'Neill]

Thank you for your help. I grasp it now.
(20 x 0.125 ) -(8 x 0.05) = 2.1 kg m. Won't topple over.