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I have been trying to come up with the optimal way to calculate the force required for an actuator to be able to lift a bridge 90 degrees. (A drawbridge) The actuator is placed as follows: (The bridge is red, its rotation point is denoted as the black dot and the actuator is pink). On the left is the starting point, and on the right - the endpoint. enter image description here

My question is how can I calculate the force required for the actuator to lift the bridge up? Also, would the force be always the same or not, as I assume once it starts lifting, the force required to continue the movement becomes smaller and smaller. What about the position of the actuator? Does its movement come into play when calculating the force or not?

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  • $\begingroup$ Movement only comes into play if you want to move the bridge fast. If you move slowly then you can approximate that static load is equal to dynamic load. You still need to find the maximum force though. $\endgroup$
    – joojaa
    May 23 '20 at 9:48
  • $\begingroup$ Depends on the counterbalance. $\endgroup$ May 23 '20 at 14:50
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enter image description here

At balance you will have the situation:

$ f\times cos(a)\times r = F \times R $

To get the bridge to lift you will need f greater than the balance value.

Note that with the bridge pivot as drawn you can't get the bridge fully up and if you lift it fully up the actuator can't pull it down as it is past "top dead-centre". Top dead-centre will occur when the line of the actuator passes through the bridge pivot.

as I assume once it starts lifting, the force required to continue the movement becomes smaller and smaller.

Correct until you start to approach top dead-centre at which point you can push as hard as you like and will get no more rotation.

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  • $\begingroup$ @Georgelua and check out why on real bridges they have massive counterweights. $\endgroup$
    – Solar Mike
    May 23 '20 at 13:00
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    $\begingroup$ The right side of the equation is only correct when the bridge is in starting position. since $r$ also depends on $\alpha$ this detail is relevant. $\endgroup$
    – mart
    May 25 '20 at 6:26

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