What Is the Acceleration and Tension in a Frictionless Inclined Plane System?

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SUMMARY

The discussion focuses on calculating the acceleration and tension in a frictionless inclined plane system involving two masses (m1 and m2) of 1.0 kg each and an angle of inclination (θ) of 30°. For part A, the correct approach involves applying Newton's second law (ƩF=ma) to derive the equations for both masses. The participant initially attempted to use the quadratic formula but was advised to reassess the forces acting on the system instead. The solution requires understanding the relationship between tension and gravitational forces without resorting to complex formulas.

PREREQUISITES
  • Understanding of Newton's second law (ƩF=ma)
  • Basic knowledge of forces on inclined planes
  • Familiarity with trigonometric functions (sine and cosine)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the dynamics of systems involving inclined planes and tension forces
  • Learn how to derive equations of motion for connected masses
  • Explore the effects of friction on inclined planes
  • Investigate the use of free-body diagrams in solving mechanics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to inclined planes and tension in connected systems.

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Homework Statement


If m1=m2=1.0 kg and θ=30°, what will be the acceleration of the system? Calculate the tension in the cord, if a block m1 lying on a frictionless inclined plane and is connected to a mass less cord. Assume m1 is moving down the plane for A), and for B) m1 is in the opposite direction. C) If m1= 1 kg and θ=30°, and the system remains at rest, what must the mass of m2 be? Calculate the tension.

Homework Equations



ƩF=ma

The Attempt at a Solution



For part A
I have this so far
m1 x: ƩF=mg(sin30°)-T=ma
T=mg(sin 30°)/ma
y: ƩF=N-mg(cos 30°)=0
N=mg(cos 30°)
m2 y: T-mg=ma
T=ma+mg
T1=T2
therefore, ma+mg=(g(sin 30°))/a
so to solve for a I end up getting
a2+ ga - gsin30°=0
so do I just do the quadratic formula here or am I going the wrong way about it?

Here's the problem if you need to see the picture

http://imageshack.us/f/812/ue4y.jpg/


 
Physics news on Phys.org
No need for the quadratic formula. look at your Forces in the direction. You made a mistake.
 

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