Helicopter carrying a water balloon

[ThomasMagnus]

Homework Statement

A helicopter accelerates horizontally while carrying a 220kg water balloon by a rope.

a)Draw a free-body diagram of the water balloon. (ignore air resistance.)

b) Find the magnitude and direction of the helicopter's acceleration.

The Attempt at a Solution

a)

b) To find the horizontal acceleration of the helicopter, we must find the sum of the horizontal forces acting on it.

The horizontal force acting on the balloon is the 'force parallel' and is:
(mg)(tan27)=(220)(9.8)(tan27)=1098.5N
$$\Sigma$$Horizontal forces on ball=ma
1098.5=220a
a=5.00m/s² to the rightThis is the correct answer from the book, but I have a few questions about it. Is the method I used correct? What about the tension in the rope? Doesn't it also pull horizontally? My method is assuming that the acceleration of the ball will be the same as the acceleration of the helicopter; is this true? With the angle of 27 degrees, I was able to find a force that acts along the x-axis (I guess it would be called the 'force parallel') Is this the same as the force applied by the plane to move horizontally?

Thanks

 

[PhanthomJay]

Homework Statement

A helicopter accelerates horizontally while carrying a 220kg water balloon by a rope.

a)Draw a free-body diagram of the water balloon. (ignore air resistance.)

b) Find the magnitude and direction of the helicopter's acceleration.

The Attempt at a Solution

a)

b) To find the horizontal acceleration of the helicopter, we must find the sum of the horizontal forces acting on it.

The horizontal force acting on the balloon is the 'force parallel' and is:
(mg)(tan27)=(220)(9.8)(tan27)=1098.5N
$$\Sigma$$Horizontal forces on ball=ma
1098.5=220a
a=5.00m/s² to the right


This is the correct answer from the book, but I have a few questions about it. Is the method I used correct?

well, not quite...right answer for wrong reason

What about the tension in the rope? Doesn't it also pull horizontally?

yes, it provides the horizonatl force...the only horizontal force acting on the balloon.

My method is assuming that the acceleration of the ball will be the same as the acceleration of the helicopter; is this true?

yes

With the angle of 27 degrees, I was able to find a force that acts along the x-axis (I guess it would be called the 'force parallel')

you can't have a force acting on an object unless it is a contact force (or an action at a distance force like gravity).

Is this the same as the force applied by the plane to move horizontally?

. The force to move the chopper is much greater than the force to move the water balloon.

Thanks[/QUOTE]

 

[ThomasMagnus]

Thank you for your response.

How is it possible to find the horizontal force of the tension in the rope by using the balloon's force of gravity?

Or what other way is there to find it?

 

[ThomasMagnus]

Is it because the tension force is composed of a horizontal force and a vertical force, and the only vertical force is the force due to gravity?

 

[PhanthomJay]

Is it because the tension force is composed of a horizontal force and a vertical force, and the only vertical force is the force due to gravity?

Yes. The vertical component of the tension force on the balloon, balances the gravity force of the Earth on the balloon, per Newton 1, and the horizontal component of the tension force on the balloon, which is the net unbalanced force in the x direction, causes the horizontal acceleration of the balloon, per Newton 2, in the direction of the net force.