Calculating Lift Force and Air Resistance in a Moving Helicopter

In summary, the conversation discusses a helicopter moving horizontally at a constant velocity and the forces acting on it, including weight, lift, and air resistance. The magnitude of the lift force is found using trigonometric equations, and the magnitude of the air resistance is determined by setting the sum of all forces in the x direction to 0. The individual seeking help eventually figures out the solution on their own.
  • #1
PhysicFailure
7
0
A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 57600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

(a) What is the magnitude of the lift force in N?

(b) Determine the magnitude of the air resistance R that opposes the motion.


I used [tex]\Sigma[/tex]Fx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)

and [tex]\Sigma[/tex]Fy= may-->Lcos21-57600


now I don't know what to do. What did i do wrong?
 
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  • #2
Hi PhysicsFailure!

I am a new member and somewhat new to physics compared to most here but I think I can help you out on this one.

First, we know that the helicopter is not accelerating in either the X OR Y directions (it is moving strictly horizontally). For this reason, the sum of all forces in each direction is 0, not just X.

Have you tried drawing a simple diagram, with the helicopter represented as a point at the origin, the weight acting downward, and the force of lift acting upward at an angle of 21 degrees to the Y axis?

I think if you set the Y equation you already have to 0, the force of lift will become apparent. From here, figure what component of that lift on the X axis must be matched by the air resistance for the copter to not have any acceleration (sum of X forces=0)

Let me know if you get it!
 
  • #3
So far so good! It looks as though you've done everything right.

"I used [tex]\Sigma[/tex]Fx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)"

So then what is the force due to air resistance?

"and [tex]\Sigma[/tex]Fy= may-->Lcos21-57600"

Part a is asking for L. How would you solve for L in this case?

Keep in mind that because there is no vertical acceleration, the net force in the y direction will also be zero.

PhysicFailure said:
A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 57600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

(a) What is the magnitude of the lift force in N?

(b) Determine the magnitude of the air resistance R that opposes the motion.I used [tex]\Sigma[/tex]Fx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)

and [tex]\Sigma[/tex]Fy= may-->Lcos21-57600now I don't know what to do. What did i do wrong?

Homework Statement


Homework Equations


The Attempt at a Solution

 
  • #4
haha thanks so much bchandler and mattowander. I actually figured out what I was not doing. I could just solve it! and for some reason i was just not thinking...THANK YOU!
 

Related to Calculating Lift Force and Air Resistance in a Moving Helicopter

1. What is Newton's 2nd Law of Motion?

Newton's 2nd Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This means that the greater the force acting on an object, the greater its acceleration will be, and the more massive the object, the less it will accelerate.

2. How is Newton's 2nd Law related to force and acceleration?

Newton's 2nd Law is directly related to force and acceleration because it explains how these two quantities are linked. The law states that the force acting on an object is directly proportional to its acceleration. This means that if the force increases, the acceleration will also increase.

3. What is the formula for Newton's 2nd Law?

The formula for Newton's 2nd Law is F=ma, where F is the net force acting on the object, m is the mass of the object, and a is the acceleration of the object. This formula shows the direct relationship between force and acceleration, with mass acting as the constant.

4. Why is Newton's 2nd Law sometimes confusing?

Newton's 2nd Law can be confusing because it involves mathematical calculations and understanding the relationship between force, mass, and acceleration. It can also be challenging to visualize how these quantities are linked in real-life situations. Additionally, some people may struggle with the concept of inverse proportionality.

5. How can I apply Newton's 2nd Law in everyday life?

Newton's 2nd Law can be applied in everyday life in various ways, such as understanding how to push or pull objects with different masses, calculating the force needed to accelerate a car, or explaining the motion of a ball when kicked. It is also essential to understand this law when participating in sports or other physical activities.

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