Why Does Using T=ma Instead of F=ma Affect the Solution in Dynamics Problems?

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SUMMARY

The discussion clarifies the distinction between using T=ma (tension) and F=ma (force) in dynamics problems. Tension (T) is specifically relevant in systems involving strings or pulleys, while force (F) represents the overall force acting on an object. The participant successfully calculated acceleration using F=ma and confirmed the book's answer. The conversation also addresses the interpretation of negative acceleration, emphasizing that it depends on the chosen coordinate system.

PREREQUISITES
  • Understanding of Newton's Second Law: F=ma
  • Knowledge of tension in physics, particularly in systems with strings or pulleys
  • Familiarity with gravitational force and its effects on mass
  • Basic principles of coordinate systems in physics
NEXT STEPS
  • Study the role of tension in pulley systems and its mathematical representation
  • Explore the concept of negative acceleration and its implications in various coordinate systems
  • Review examples of dynamics problems involving both tension and force
  • Learn about the effects of friction on motion and how it alters acceleration calculations
USEFUL FOR

Students studying physics, particularly those focusing on dynamics, mechanics, and the application of Newton's laws. This discussion is also beneficial for educators seeking to clarify common misconceptions in force and tension concepts.

Abdilatif
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Homework Statement


http://myclass.peelschools.org/sec/11/22607/Lessons/Unit%202%20Dynamics/Review%20Chapter%202.pdf Question 12, a and b?

Homework Equations


F=ma , Force = mass* acceleration
a=g(m2-usm2)/(m1+m2), g=gravity, us=coefficient of static friction, m1=mass1, m2=mass2

The Attempt at a Solution


a=F/m so the mass would be 6Kg, if i use the mass of the whole system. The force, would be only for the mass being acted on, which is F=mg, so it is F=2*9.8=19.6N. That is the force acting on the system. So back to the first equation, F=ma, 19.6=6a, simple math and a=3.3m/s. The book has the same answer.

Though I am puzzled to why they use T=ma instead of using F=ma, what does T stand for? Also by doing some more math, you know that the general equation for a = g(m2-usm2) / m1+m2, g=gravity, us=coefficient of static friction. So, then when I plug in my values I get a negative value is that right? Either way, in which cases are acceleration negative, when accelerating backwards, or when falling?
 
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Just to alert you, your link is broken, but if you have the same answer as the book, you're fine.

You asked why the book used T instead of F. T stands for tension, which is a type of force that seeks to pull something apart; basically, it can be found in strings or pulleys, which I'm guessing this problem is about.

As for the negative vales for acceleration, it depends on how you assign your coordinate system. Most commonly, the left (negative x) direction is negative, and the down (negative y) direction is negative also, leaving the right (positive x) and up (positive y) directions to refer to positive values. Let's say you have a system accelerating downwards toward the ground under the influence of gravity, and a resistive force of some kind (maybe air friction) acts on it to slow its acceleration. Furthermore, let's use the sign convention I mentioned before, where down is negative and up is positive. In this case, the value of g would be considered negative, whereas the value of acceleration the air friction causes is considered positive. If the system is still accelerating downward even when air friction acts on it, then it is moving in the downward direction (the same direction as g) and is thus considered negative. So receiving a negative value for your answer doesn't mean it's wrong; it just all depends on which directions you call negative and which you call positive. So in the general equation you gave, if you decided g was negative and the system was still accelerating in the direction of g, then the system's acceleration was also negative.
 
T stands for Tension which is given by the downward component mg
 

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