Calculating Lift Force for Semi-Circle Moving Sideways

[cmorency]

What's the equation for the force of a lift. Drag formula is 1/2 ρ∆v^2 C A. And if you need the constant how would i determine the lift constant for a semi circle shape which spins sideways and has an acceleration of approximately 8.5 m/s down (should be gravity which is 9.8 ms/s). It is moving forward at an initial speed of 62.9 m/s if that helps.

 

[Astronuc]

Is the downward acceleration = 8.5 m/s²?

Lift is a force. If an object is in free fall without an opposing force, it would accelerate at g, assuming it's near sea level. If the object is accelerating downward with a rate less than 9.8 m/s², then there must be an opposing force - drag or lift.

What is the significance of the diffrence between g and 8.5 m/s²?

 

[cmorency]

Because one shape had a gravitational pull or downward acceleration of 11.5 m/s and one 8.5 and I'm attempting to calculate the force on both objects and why.

 

[RoyalCat]

Try applying Newton's Second Law to the falling object.
You know that it is acted on by two forces, gravity and the lift force. Draw a free body diagram, and remember that [tex]\Sigma \vec F = m\vec a[/tex], the sum of all forces on a body is equal its mass times its acceleration. Knowing the acceleration, you can find what you're looking for.

 

[rdl]

The equation for lift force can be represented as F = 1/2 ρ∆v^2 S, where ρ is the density of the fluid, ∆v is the change in velocity, and S is the surface area of the object. This equation is known as the lift equation and is commonly used in aerodynamics.

To determine the lift constant for a semi-circle shape moving sideways, we would need to take into account the angle of attack (the angle at which the object is moving relative to the direction of the fluid flow), the shape and size of the semi-circle, and the fluid properties. The lift constant can be calculated using experimental data or through simulations using computational fluid dynamics (CFD) software.

In the given scenario, the acceleration of 8.5 m/s^2 downwards can be considered as the change in velocity (∆v) and the initial speed of 62.9 m/s can be used as the reference velocity (V) in the lift equation. The lift force can then be calculated as F = 1/2 ρ(V^2 - ∆v^2) S.

It is important to note that the lift force generated by a semi-circle shape moving sideways will be significantly less compared to a similar shape moving in a straight line, due to the presence of drag forces. Therefore, the lift force may not be sufficient to counteract the force of gravity in this scenario.