PHYSICS:  1.  Two students are standing on skateboards facing each other at rest. Student A (mass of 50.0 kg) pushes on student B (mass of 60.0 kg) with a force of 150 N [W]. You may assume there...

PHYSICS: 

1.  Two students are standing on skateboards facing each other at rest. Student A (mass of 50.0 kg) pushes on student B (mass of 60.0 kg) with a force of 150 N [W]. You may assume there is no friction.

a) Draw the FBD of each skateboarder. 

b) Find the acceleration of each skateboarder. 

2. A helicopter of mass 4500 kg is hovering at rest above the ground. 

a) Draw an FBD of the hovering helicopter.

b) Calculate the forces acting on the helicopter while it is hovering.

c) Explain what causes the upward force on the helicopter.

d) Find the acceleration of the helicopter if the propeller blades apply a force of 49 000 N [down] on the air. 

Asked on by deni123

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sciftw's profile pic

sciftw | High School Teacher | (Level 1) Educator Emeritus

Posted on

I am sorry, but I don't know how to effectively draw something using this platform. I can answer the other parts of your question though. 

The acelebration of each skateboarder can be found using the formula acceleration = force/mass. Newton's third law explains that the force on each skateboarder is the same size. You do need to use the equation twice though because the two skaters have different masses. 

Student A:  A = F/M    A = 150/50.    A = 3 m/s/s 

Student B: A = F/M     A = 150/60     A = 2.5 m/s/s

2b. Calculate the forces acting on the helicopter while it is hovering. 

That isn't 100% possible with the information given. I can calculate the upward and downward force, but there is likely some crosswinds acting on the aircraft as well. What I do know for sure is that the net of all of the forces acting on the helicopter is zero. That is because the helicopter is hovering, which means that it is not moving. In other words there is no change in its motion or velocity.  That means balanced forces, which have a net force of zero. Knowing that allows us to say that whatever the downward pull of gravity is, the upward lift of the blades is equal in size. 

Use the formula A = F/M but rework it to isolate F. Then the formula is F = MA. You know the mass, and acceleration in this case is the acceleration due to gravity,  which is 9.8 m/s/s. F = 4500 x 9.8    F = 44,100 N. So gravity is pulling down with a force of 44,100 Newton's and the blades provide exactly that much lift in order to keep the helicopter hovering. 

2c: The upward force is generated by the spinning rotor blades.  They generate lift. Each blade is shaped like a wing. Flat on bottom and curved on top. As the blade moves through the air, the air over the top of the blade moves faster. Bernoulli's principle states that a faster moving fluid exerts less pressure. The spinning blades create an air pressure difference.  Low pressure above the blades, and high pressure below the the blades. In order to equalize, the high pressure air tries to move to the low pressure area, which lifts up on the helicopter blades; therefore lifting the entire helicopter.  

2d. Use A = F/M.  A = 49,000/4500.   A = 10.89 m/s/s upward. That would be its acceleration, if gravity wasn't active. I'm assuming gravity is still active, so we need to use the helicopter's "resultant force." That is the upward force of its blades minus it's weight. 49,000 - 44,100 = 4,900.  

A = 4,900/4,500.      A = 1.089 m/s/s upward. 

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