# How to calculate weight and force required to lift a flap?

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1. Jun 29, 2015

### OJT

Hi all,

I would like to know how to calculate the force required to lift a flap (or drawbridge if you proffer) of known weight that is on a pivot and where the pull chain (let’s say it’s a chain) connects to the flap at a known distance from the pivot.

I’d also like to be able to calculate the effective weight it would exert when resting at it’s top (denoted by a blue dot on the drawing)

I’m no engineer or mathematicians but I do have CAD softwear so can measure angles (as shown on the attached drawing) so the simplest methods - i.e. without any trigonometry if possible - would be great! I have been unable to find similar explanations online with equations that the person un-versed in long equations can follow that easily(!!) – although am happy to learn if there is no simple solution!!

So the flap / drawbridge:

- Is 3.2m long

- Has a weight of 25kg.

- Pivots around the point marked ‘p

- The ‘chain’ connects to the flap at a distance of 2m from the pivot / or 1.2m from the free end

- All (I hope) relevant angles are shown on the drawing.

The problems to solve:

1. How can you calculate the force required to lift the flap from horizontal to 45 degrees (A) and from 45 degrees to 12 degrees (B)?

2. How would you calculate the weight exerted on the resting point at the top of the flap (as denoted by the blue dot). 'C' denotes the direction I think this force would be acting in.

Any help or pointers will be fantastic so thanks in advance,

Oliver

#### Attached Files:

• ###### Effective weight.pdf
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2. Jun 29, 2015

### BiGyElLoWhAt

Well you need to actually look at Torque. T=RxF (or R*F*sin(angle in between)) and sin(angle) will effectively be 1 assuming you build it so the rope makes a 90 degree angle with the flap. The weight of the rock needs to be able to overcome any weight torques acting on the bridge (which will not always be 90 degrees). However, the Torque of gravity on the bridge will only decrease as you rotate the bridge upwards, so as long as the bridge torque (which, by the way acts through the center of mass) while the bridge is down, it will be sufficient (assuming no friction on the pulley).