Tension of 2 Bodies on a Frictionless Inclined Plane

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Homework Help Overview

The problem involves two boxes of different masses on frictionless inclined planes connected by a cord over a frictionless pulley. The original poster attempts to calculate the tension in the cord given the masses and angles of inclination.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law to each mass and question the use of negative signs in the equations. There is also a focus on the correct angles used in the calculations.

Discussion Status

Some participants have offered observations about potential errors in the original poster's calculations, particularly regarding the signs in the equations and the angles used. Multiple interpretations of the equations and their setup are being explored.

Contextual Notes

There is mention of significant figures in the results, and participants are addressing possible errors in the transcription of the angle for one of the masses.

ewiner
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Homework Statement



Figure 5-60 shows a box of dirty money (mass m1 = 2.7 kg) on a frictionless plane inclined at angle θ1 = 28°. The box is connected via a cord of negligible mass to a box of laundered money (mass m2 = 2.4 kg) on a frictionless plane inclined at angle θ2 = 64°. The pulley is frictionless and has negligible mass. What is the tension in the cord?

Homework Equations



F = ma

T - mgsin\Theta = -ma

a = ( m / M + m )g

The Attempt at a Solution



Newton's 2nd law for each body:
m1: T - (2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)a
m2: (2.4 kg)(9.8 m/s2)sin68 - T = -(2.4 kg)a

Add m1 & m2: (2.4 kg)(9.8 m/s2)sin68 - (2.7 kg)(9.8 m/s2)sin28 = (2.4 + 2.7)a

Solve for a: a = 1.84

Insert a into m1: T-(2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)(1.84 m/s2)

T = 17.4 N

However, that is not the right answer... Where am I going wrong?
 

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Why do you have a negative sign in front of the masses? If the acceleration is negative it'll come out in your answer when you solve for a. You might be counting the negatives twice. Just my 2 cents.
 
Although you incorrectly threw in those negatives in front of the masses, you ignored that error and changed it correctly to a positive number when you did your equation additions and substitution to solve for T. Strange. But the other issue is that you copied the angle down incorrectly...the angle given is 64 degrees, but you used 68 degrees. Make that angle correction and solve for T using 2 significant figures in the result.
 
PhanthomJay said:
Although you incorrectly threw in those negatives in front of the masses, you ignored that error and changed it correctly to a positive number when you did your equation additions and substitution to solve for T. Strange.

Then he went back to the negative

Insert a into m1: T-(2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)(1.84 m/s2)
 

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