Lever question when lever is angled

[JCaleb]

Hello, I have a question about a see-saw / lever problem. Does this configuration I have shown here allow me to treat the system like a regular lever if I make the downward force at the horizontal point from the center to be the cosine of the angle? In this example the downward force on the left would be 2kg at 1m whereas on the right it would be 1kg * cos(5) at 2m.

Based on the static equilibrium equation of a lever F1*d1= F2*d2 the system would not be in equilibrium and would fall to the left. 2*1 > 0.64*2

Is this correct?

Would that mean that there is a compression force of the sine of the angle down the shaft towards the fulcrum on the right side? 1kg*sin(50) ?

 

[dE_logics]

In this example the downward force on the left would be 2kg at 1m whereas on the right it would be 1kg * cos(5) at 2m.

How do you define the distance?

Based on the static equilibrium equation of a lever F1*d1= F2*d2 the system would not be in equilibrium and would fall to the left. 2*1 > 0.64*2p

The object will fall unless a 1m point is made to hold. If this is done, then it'll be a net equilibrium.

No it will not rotate while falling due to gravity.

 

[astrokreng]

Yes, your understanding is correct. In this configuration, the lever would not be in equilibrium and would fall to the left due to the imbalance of forces. The compression force on the right side would be equal to the sine of the angle multiplied by the downward force of 1kg. This is because the downward force is not acting directly perpendicular to the lever, but rather at an angle, creating a component force in the direction of the lever. This is an important consideration in lever systems and must be taken into account when calculating equilibrium. I would also recommend using the cosine of the angle on the right side, as it is the adjacent side in this scenario. Overall, your analysis is correct and demonstrates a good understanding of lever mechanics.