Problem: Watch attached to a string in an airplane

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The discussion centers on a physics problem involving a watch dangling from a string in a jetliner during takeoff, where the string makes a 25-degree angle with the vertical. The original approach incorrectly assumes three unknowns without a proper relationship between them, leading to confusion in solving for acceleration and tension. Correctly, the vertical acceleration should be set to zero, allowing for the calculation of tension first. Once tension is determined, it can be used to find the horizontal acceleration. The key takeaway is to align the coordinate system with the direction of motion for accurate analysis.
N_L_
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I'm having trouble with the correct setup for the following problem:

A physics student, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from JFK Airport. She notices that the string makes an angle of 25 degrees with respect to the vertical as the aircraft accelerates for takeoff, which takes about 18 seconds. Estimate the takeoff speed of the aircraft.


Given: Theta = 25 degrees ; time = 18 s
Unknown: Ft, acceleration (ax and ay)

I drew an FBD and split up gravity into its x and y components.

I have the equations as:

mg sin 25 = m ax

Ft - mg cos 25 = m ay

What's wrong with this approach?
 
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Here's my solution to your problem
a=g\tan\theta
where \theta=25^o
And the takeoff speed of the aircraft is
v=at=gt\tan\theta[/color]
 
N_L_ said:
I'm having trouble with the correct setup for the following problem:

A physics student, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from JFK Airport. She notices that the string makes an angle of 25 degrees with respect to the vertical as the aircraft accelerates for takeoff, which takes about 18 seconds. Estimate the takeoff speed of the aircraft.


Given: Theta = 25 degrees ; time = 18 s
Unknown: Ft, acceleration (ax and ay)

I drew an FBD and split up gravity into its x and y components.

I have the equations as:

mg sin 25 = m ax

Ft - mg cos 25 = m ay

What's wrong with this approach?

First, what forces do you have in your FBD? There are only two forces: the tension in the string and gravity.

Second, a_y = 0.

Imposing a_y = 0 will allow you to find the tension. Then using the tension you found and plugging it into the x equation will allow you to find the acceleration along x.

The answers of the other posters are right.

Patrick
 
As nrged implied, ALWAYS set up a coordinate parallel to the direction of motion or parallel the acceleration. (implied by not even considering that anyone might set up a coordinate system parallel the string).
Otherwise, what's wrong with this approach is that you have 3 unknowns
(ax, ay, T ) so you need an equation to relate ax with ay.
By geometry, ax + ay must add up to a horizontal total acceleration,
and you're back to the "straight-forward" equation.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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