I've been trying, as a thought exercise, to run some calculations on a floor jack-like mechanism that is expected to lift 130kg (287 lbs). Dimensions and setup are as per the image below.
The 130kg load is to be positioned at the end of the boom, to simplify things, I assume the load would be at 90 degrees relative to the boom. My understanding, is that the force would be greatest then, anyway...
Here's my approach to try and size the motor: 1. Find the torque at point A, due to the load on the end of the boom.
$$F = ma = 130\cdot10 = 1300\text{ N}$$
I specify $A$ as 10, rather than 9.8. So, 130 kg load exerts a force of 1300 N.
The torque, then at point A would be
$$\tau = rF\sin\theta = 0.5 \cdot 1300 \sin(90) = 650\text{ Nm}$$
- I don't know what force the linear actuator needs to produce - x. I proceed to calculate what applied force at point B would yield a torque greater than 650 Nm. Let's assume a 10% safety margin, so 715 Nm needed at point B.
The angle will go from 90 degrees down, let's say to 30 degrees at the max lift point of the boom.
With reduced angle, the torque is also diminished, so at the highest point, the linear actuator's force applied will need to be (at 30 degree angle relative to the boom)
$$715 = rF\sin\theta = 0.1F\sin(30°) \therefore F = 23,833\text{ N}$$
This sounds like a ridiculously high force rating for an electric linear actuator.
Is my approach correct? I'm quite certain I'm missing something important. Your help will be greatly appreciated.
EDIT Clearly, the above setup isn't ideal for an electrical linear actuator. The other lifting setup I can think of, is a mobile crane, like in the pic below.
Taking the same size and parameters as above, the force needed by the actuator is much more palatable.
Let's assume force is applied at a distance of 40cm from the point of rotation and an initial 30 degree angle. The lifting mechanism needs to counter a torque of 715Nm (calculated above).
$$715 = rF\cos\theta = 0.4F\cos(30°) \therefore F = 2,102.94\text{ N}$$
Is this correct? (ignoring the counter-weights on the actual crane)