Winch rope force to pull back a folding canopy

[1988ajk]

Hi all....

Struggling a little to determine the force required on the rope to pull 2x hinged leaf panels back. the leaf panels form a folding roof that will be opened/closed by a winch.

Each link weighs around 80Kg. theta starts around 5 degrees from horizontal and ends around 75 degrees from horizontal.

Thanks.

 

[256bits]

Struggling

I see your diagram with the top pivot point between L1 an L2 as having only pin connection.
Should not then the forces on the pin and sleeve be completely horizontal, much like a frictionless roller on a vertical wall.

 

[1988ajk]

There will be vertical reactions at the top pivot point due to the masses (mg) of each leaf. We are assuming this works similar to a toggle clamp mechanism. However, I am sure there are other methods of summing the forces to find the horizontal load at the rope.

 

[Baluncore]

Maximum tension will be at 5°. Half of the mass of each leaf rests on the ground, the other half at the top pivot, making 80 kg total at the top.
Tan(5°) = 80 kg / tension kg
Tension in rope; t = 80 / tan(5°) = 915 kg.
Tensional force in rope = 9.8 * 915 kg = 8967. newton.

 

[Lnewqban]

Please, see:
https://www.engineeringtoolbox.com/toggle-joint-d_2077.html

 

[256bits]

There will be vertical reactions at the top pivot point due to the masses (mg) of each leaf

No there will not be.
A toggle joint cannot support vertical forces transmitted through the joint.
Any vertical force put at the location of the joint instead results in compression of the leaf members.

This

I see your diagram with the top pivot point between L1 an L2 as having only pin connection.
Should not then the forces on the pin and sleeve be completely horizontal, much like a frictionless roller on a vertical wall.

this

Maximum tension will be at 5°. Half of the mass of each leaf rests on the ground, the other half at the top pivot, making 80 kg total at the top.
Tan(5°) = 80 kg / tension kg
Tension in rope; t = 80 / tan(5°) = 915 kg.
Tensional force in rope = 9.8 * 915 kg = 8967. newton.

and this

Please, see:
https://www.engineeringtoolbox.com/toggle-joint-d_2077.html

are all the same thing.

 

[256bits]

It is important to pick the correct P when using the formula for toggle joint:

The answer from engineering toolbox : 457 kg tension at 5°.

With the frictionless wall method.
Taking moments about the pivot point for a single leaf:
mg /2( L1 cos ∝ ) + T( L1 sin∝) = mg (L1 cos∝ )
T = mg/( 2 tan ∝ ) = 80 / (2 tan 5° ) = 40 / .0875 = 457.1 kg tension @ 5°.

T = mg/( 2 tan ∝ )
T/g = m /( 2 tan ∝ ) = 80 / (2 tan 5° ) = 40 / .0875 = 457.1 kg tension @ 5°.
T = 457.1 kg * 9.81 m/s^2 = 4483.2 N

 

[256bits]

making 80 kg total at the top

Should be 2P/2=40 for one leaf at the top.

 

[berkeman]

457 kg tension at 5°

kN?

 

[256bits]

kN?

Edited Post #7
thanks.