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Tension of an object hanging in an airplane

  • #1

Homework Statement



On her flight home at Thanksgiving, Margaret decides to apply the physics she has learned in PHYC1300 to measure the speed of the aircraft when it takes off. She dangles her watch from a string while the aircraft accelerates down the runway. Draw a free body diagram of the watch. Using her protractor, she notices that the string makes an angle of 22° with respect to the vertical edge of the window as the aircraft accelerates on the runway, which takes about 22 s. Estimate the takeoff speed of the aircraft.

Homework Equations



[itex]\vec{F}_{net}=m\vec{a}[/itex]

[itex]\vec{w}=-mg[/itex]

(here, [itex]\vec{w}[/itex] is the force of gravity)

The Attempt at a Solution



I find it difficult to draw diagrams for problems on computers, so hopefully words will suffice. I have the acceleration going to the left, and so the Cartesian axes are in the usual directions. [itex]\vec{w}[/itex] pulls straight down on the watch. Force of tension in the string, [itex]\vec{T}[/itex], points up and to the left, making the 22-degree angle with the y-axis. Breaking tension into components, I see that the x-component can be described with sine, because it is opposite the angle, and the y-component, with cosine, since it is adjacent:

[itex]\vec{T}_{x}=-|\vec{T}|sinθ[/itex]
[itex]\vec{T}_{y}=|\vec{T}|cosθ[/itex]

Also, since the acceleration is only in the x direction, [itex]\vec{w}[/itex] and [itex]\vec{T}_{y}[/itex] must be equal and opposite. Thus

[itex]\vec{T}_{x}=m\vec{a}[/itex]

[itex]-|\vec{T}|sinθ=m\vec{a}[/itex]

My goal for the moment is to find acceleration, and attempt to find speed from there. Unfortunately, you can see that I have an equation with two unknowns (mass and tension).
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
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Also, since the acceleration is only in the x direction, [itex]\vec{w}[/itex] and [itex]\vec{T}_{y}[/itex] must be equal.
This will give you a second equation that will allow you to solve for the acceleration.
 
  • #3
This will give you a second equation that will allow you to solve for the acceleration.
Thanks, I'm not sure how I didn't see it, but just this little prompt helped me with the algebraic manipulation. In case anyone needs help with this sort of problem in the future:[tex]m=(|\vec{T}|cosθ)/g[/tex]

[itex]\vec{a}=(-g|\vec{T}|sinθ)/(|\vec{T}|cosθ)[/itex]
[itex]\vec{a}=(-gsinθ)/(cosθ)[/itex]
 

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