tipping calculation

[mir]

it will be that I am particularly rusty but I do not understand how to determine the f that makes me turn the system:
consider these values:
a = 200 mm (base age)
b = 500 mm
c = 400 mm (center height)
fp = 500 n
alpha = degree of inclination (graphically it is 26.57°)


How much is f?

Thanks for the help!
♪

 

[meccanicamg]

In your case fp I imagine it is the weight strength of the structure and it is already flush of the base. therefore just a breath of wind to bring the force out of 0.01 mm

 

[mir]

we say that the balance is fp x a = f x b

so until "f x b < fp x a" does not turn

 

[Stan9411]

No. for how you drew fp has no arm compared to the rotation center so it does not participate in the balance of moments. the f arm instead is the vertical distance between f and the base of the structure on the ground, that is 536,65 . strength and arm must always be perpendicular

 

[PiE81]

therefore, as mentioned by mechanicsmg, for any positive f value (concord with the drawn vector) the structure turns over.

 

[mir]

No... you must violate the balance that I said above.

 

[Stan9411]

I think you're wrong. the center of rotation is the point where the base touches ground and not the one where rod and base meet. in this specific fp configuration is aligned vertically with the center of rotation so it does not participate in balance.

 

[mir]

 

[PiE81]

Well, it seems to me that the concept is clear to you. Since in the initial image k=0, the equation m=f*h-fp*k is reduced to m=fh, confirming the above.

 

[Stan9411]

I was going to write it down

 

[PiE81]

and anyway, placing alpha = 0, with the condition you wrote you are looking for the f force that makes you lift the system from the ground, do not tilt.