# How to calculate force / torque on non-flat lever, i.e. dolly

See attached image. The mass is being rotated on a lever where the pivot point (P) is a certain distance ($L_2$) from the right angle at the bottom. How do I calculate the force necessary to apply horizontally at point U to lift the mass in the worst case (i.e. where the rotation position requires the maximum force), ignoring for now the weight of the lever itself, friction, etc. The structure will never rotate counter-clockwise from its illustrated position, and will rotate up to $60^\circ$ clockwise. Also assume the mass will be distributed evenly across $L_1$, i.e. the center of the mass is in the center of $L_1$.

For those interested: this is part of a robotics project. A string attached to a motor/pulley system will be pulling at point U. I'm trying to determine if the motor has sufficient stall torque and if so, how much mass we can reasonably expect to be able to lift.

You need to balance the moments about point P. The horizontal force F time $L_3$ will equal the mass $M$ times the horizontal distance $L_4$ between P and M.

$$F \times L_3 = M \times L_4$$

This calculates the force required in the current position you've shown.

Worst-case, an infinite force will be required after rotating $90^{\circ}$ from that shown, assuming that the mass M is a point mass located on the lower bar.

$$F \times 0 = M \times L_2$$

If the lift angle is $\theta$ (shown at zero in the diagram) then the payload lever arm is $$x_1 = \tfrac{L_1}{2} \cos \theta+L_2 \sin\theta$$

The force lever arm is $$x_3 = L_3 \cos\theta$$

Static balance exists when $$\left. \vphantom{\int } (M g) x_1 = F x_3 \right\} \\F = \frac{x_1}{x_3} M g = \frac{\tfrac{L_1}{2} \cos \theta+L_2 \sin\theta}{L_3 \cos\theta} M g$$