I'm developing a prop and need to figure out how much force will be required to lift/push an anchored lever. Here is a simple drawing.
The idea being that the slab is hinged on one side. There is a small rod which is attached to the slab also on a hinge. There is a linear force pushing the unhinged side of the rod towards the hinged base of the slab. The slab is intended to raise. The gray image shows the position of the slab and lever after it's been raised some. Obviously with the setup shown in the image it will not raise past a certain point, but it's a good basis for the calculations I need.
It seems obvious that the force will change as the slab raises (more force the lower/more horizontal the slab is). All I really need is the max force. (anything less is obviously covered). All the weights, angles and measurements will be known. I'm not looking for the answer, but rather for the math/formulas so I can figure out where exactly to place the lift rods, the length of the rods, etc. I don't want to start buying parts only to find the necessary force will be way too high to do it like this.
Best Answer
Consider all movable members of the prop to be immobile. We can then invoke Newton's law; the sum of vertical ($y$) forces on each member and the sum of horizontal forces ($x$) must be zero: $$\Sigma F_y=0$$ $$\Sigma F_x=0$$
1. Applied to the plate:
$$F_3\sin\beta+F_2\sin\alpha=mg$$ $$F_2\cos\alpha-F_3\cos\beta=0$$ Or: $$F_2\sin\alpha=mg-F_3\sin\beta$$ $$F_2\cos\alpha=F_3\cos\beta$$ Or: $$\tan\alpha=\frac{mg}{F_3\cos\beta}-\tan\beta$$
2. Applied to the rod $AP$:
$$F_3\sin\beta+F_1\sin\beta=0 \implies F_3=-F_1$$ $$F_1\cos\beta=F_3\cos\beta+F$$ $$\implies F_3=-\frac{F}{2\cos\beta}$$ Combined with: $$\tan\alpha=\frac{mg}{F_3\cos\beta}-\tan\beta$$ We get: $$\large{F=\frac{2mg}{\tan\beta-\tan\alpha}}$$ If $F$ exceeds this value the plate will move upwards (rotate about $A$).
Note that for a horizontal plate: $\alpha=\beta=0\implies F\to+\infty$.