Tension in Strings: Solving a .5kg Problem

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Homework Help Overview

The problem involves determining the tension in two strings supporting a 0.5 kg mass, with the longer string measuring 0.5 m and positioned at specific angles. The original poster outlines their approach by breaking down the forces into x and y components and performing calculations based on equilibrium conditions.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to resolve the tension in the strings by applying equilibrium equations and trigonometric relationships. Some participants question the choice of gravitational acceleration, discussing the appropriateness of using 10 m/s² versus 9.8 m/s² in calculations.

Discussion Status

Participants have acknowledged the original poster's method as valid, with some providing insights into the conventions used in AP Physics regarding gravitational acceleration and trigonometric calculations. There is an ongoing exploration of assumptions related to the values used for g and the angles involved.

Contextual Notes

There is a mention of the problem being part of a series, which may influence the choice of gravitational acceleration. Participants note that certain angles are favored in non-calculator sections of exams, suggesting constraints on the types of trigonometric functions that may be relevant.

aeroengphys
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For homework, I was given the following problem:

A .5kg hangs from two strings at the angles shown. The longer string is .5m long.
(a) Determine the tension in each string.

(see attachment for diagram)Can you tell me how this looks...

1st I broke everything into x and y components (ie Tlx, Tly, Trx, Try) then i did the following:

x
ΣF=0N
Tlx=Trx
Tlcos(60)=Trcos(25)
Tl=Trcos(25)/cos(60) y
ΣF=0N
Tly + Try = 5N
Tlsin(60)+Trsin(25) = 5N
(Trcos25/cos60)*sin60 + Trsin25= 5N
Tr((cos25*sin60)/cos60) + sin25= 5N
Tr = (5N - sin25)/((cos25*sin60)/cos60)
Tr = 2.92NThen...I went back to my x components and plugged in Tr and solved for Tl as shown:

Tl = (2.92N)(cos25)/cos(60)
Tl = 5.29N

Hopefully that's right...Thanks in advance for letting me know.
 

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I didn't check your arithmetic, but your method is perfect. (Are you taking g = 10 m/s^2? Often 9.8 m/s^2 is used.)
 
In AP Physics B, they tell you to use 10m/s² since on the multiple choice you won't be able to use a calculator.
 
aeroengphys said:
In AP Physics B, they tell you to use 10m/s² since on the multiple choice you won't be able to use a calculator.
That's cool. As long as you are doing so on purpose. :smile:

Do they really expect you to calculate trig functions--like cos(25)--without a calculator?
 
calculations involving g will undoubtedly appear on the non-calculator paper.Questions involving sin25 are highly unlikely to.

sin/cos of 0, 30, 45, 60 and 90 maybe
 
This is a part II question, so technically you could probably use 9.8, but my teacher told us to just keep using 10. As for trig functions, they use sin/cos of 30/45/60/90 on the AP exam.
 
the AP equation sheet includes sin/cos/tan of selected angles (0,30,45,60,90).
They don't use other angles in the non-calculator section
 
aeroengphys said:
This is a part II question, so technically you could probably use 9.8, but my teacher told us to just keep using 10.
They couldn't demand you use different values on different papers - that's just asking for confusion.
 
It's always good to use 10 m/s^2 anyway for quick checks (eg multiple choice) because it gives accuracy up to two places.
 

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