the Lever

>>8012196
I think you should investigate rotational effects- for example, torque, before applying N2L in x and y to get the amount of force that must act perpendicularly to the arm; that makes finding your F_b for any angle beta pretty trivial. Once you know that, just solve for the forces in the x directions and you're good.

>>8012213
Oh wait, just saw the second calculation.
No, that's fucked.
F_a is
[math]
|\vec{Q}|sin(\alpha)
[/math]
and F_b is
[math]
|\vec{Q}|cos(\alpha)
[/math]
And then you can apply the "law of the lever" (really just newton's second law for rotational motion applied to a static system), but that only can tell you how much force is acting perpendicularly to the lever arm.Those vectors that you have at the end do not sum to Q

>>8012213
I have broken down the load Q into two forces acting along the hydraulic actuators - it is the only means of moving the crane arm, so the forces have to be calculated along those axes.

After having calculated Fa and Fb, I already have the force that acts along the arm. However, I still need to reapply the force Fb so that it aligns with the second hydraulic actuator.

I was under the impression that it boils down to equating the moment Fb*L*sin(specific_angle) and Fb' * l * sin(specific_angle). As the angles are the same, it would reduce to standard lever advantage.

Is there anything wrong with the approach?

>>8012228
Those forces are wrong.

File: problem2.jpg (211KB, 1325x1179px) Image search: [Google]

211KB, 1325x1179px

>>8012217
1) I 'need' to calculate the force going along the angle beta. A force perpendicular to the arm is of no use.

2) F_a and F_b sum geometrically to Q, I've checked. If I move F_b to F_b', the reaction force will be different (greater) - isn't that to be expected?

Consider the attached image. If the arm was supported at its end, F_b would stay as it is. However, in the original problem it is not so.

>>8012231
I can't see the fault, other than that the forces should point in the opposite directions.

File: 1460939850779-2047579334.jpg (3MB, 3984x2988px) Image search: [Google]

3MB, 3984x2988px

>>8012265
1) you just don't know how torque works
2) solved

>>8012314
I am analyzing this. Which angle is theta?

>>8012388
alpha. Instinctively wrote theta.

>>8012391
I've done calculations. My F_b' and your F_b' are completely identical.

However, your F_a is much bigger. I'm looking into it.

>>8012433
It really shouldn't be.
[math]
F_x = F_a + F_b'cos(\beta - \alpha) - Qsin(\alpha) = 0
[/math]