What Is the Plane's Constant Acceleration During Takeoff?

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Homework Help Overview

The problem involves an airplane's takeoff sequence, focusing on its constant acceleration and the forces acting on a pocketwatch held by a passenger. The context includes analyzing the forces in relation to the angle of the watch with the vertical during takeoff.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the lack of a given mass and whether it is necessary for solving the problem. There are attempts to analyze the forces acting on the pocketwatch, including tension and weight, and to break these forces into horizontal and vertical components.

Discussion Status

The discussion is ongoing, with participants exploring the components of forces acting on the pocketwatch and questioning their understanding of how to apply Newton's laws. Some guidance has been provided regarding the identification of forces and their components, but clarity on the application of these concepts is still being sought.

Contextual Notes

Participants are working under the constraint of not having a specific mass value and are questioning how to proceed without it. The angle of the watch and the time before takeoff are also key elements under discussion.

asheik234
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Frensley_Forces_Horizontal_006.gif
1. Homework Statement
An airplane begins its takeoff sequence moving with a constant acceleration a. A passenger holds up a pocketwatch during the takeoff sequence and notices that the watch makes an angle θ = 14° with the vertical, and that 13.3 seconds pass before the plane leaves the runway.

(a) What is the plane's constant acceleration?(b) How far does the plane travel on the runway?2. Homework Equations

F = ma

3. The Attempt at a Solution

There is no given mass, I don't know how to find it without it.
 
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asheik234 said:
There is no given mass, I don't know how to find it without it.
Just call the mass 'm'. You won't need the actual value.
 
Doc Al said:
Just call the mass 'm'. You won't need the actual value.

can you explain how?
 
asheik234 said:
can you explain how?
Apply Newton's 2nd law. What forces act on the pocketwatch? Analyze horizontal and vertical force components separately.
 
Doc Al said:
Apply Newton's 2nd law. What forces act on the pocketwatch? Analyze horizontal and vertical force components separately.

I tried but still nothing, the vertical component is mg and the horizontal one is 9.5m(mg*cos14), am I missing something?
 
asheik234 said:
I tried but still nothing, the vertical component is mg and the horizontal one is 9.5m(mg*cos14), am I missing something?
First things first. What forces act on the watch? (There are two forces acting: Name them.)
 
Doc Al said:
First things first. What forces act on the watch? (There are two forces acting: Name them.)

Force tension and force weight
 
asheik234 said:
Force tension and force weight
Excellent. What are the horizontal and vertical components of each force? (The weight is mg; call the tension force "T".)
 
Doc Al said:
Excellent. What are the horizontal and vertical components of each force? (The weight is mg; call the tension force "T".)

Components of force weight:
horizontal component: 0
vertical component: mg

Components of force tension:
horizontal component: mg*cos14
vertical component: mg
 
  • #10
asheik234 said:
Components of force weight:
horizontal component: 0
vertical component: mg
Good. (The vertical component acts down.)

Components of force tension:
horizontal component: mg*cos14
vertical component: mg
No. (The vertical component will end up equal to mg, but you'll get to that in the next step.)

The tension force is T. It acts parallel to the chain of the watch. What are its components?
 
  • #11
Doc Al said:
Good. (The vertical component acts down.)


No. (The vertical component will end up equal to mg, but you'll get to that in the next step.)

The tension force is T. It acts parallel to the chain of the watch. What are its components?


horizontal component: T * sin14
vertical component: mg
 
  • #12
asheik234 said:
horizontal component: T * sin14
vertical component: mg
Good.

But I would have preferred:
Horizontal: T*sin14
Vertical: T*cos14

Nice and simple.

Now apply ƩF = ma to the vertical and horizontal components.
 

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