# Help on a dynamics/kinematics problem.

1. Jan 22, 2012

### Abdilatif

1. The problem statement, all variables and given/known data
http://myclass.peelschools.org/sec/11/22607/Lessons/Unit 2 Dynamics/Review Chapter 2.pdf Question 12, a and b?

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

3. 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?

2. Nov 11, 2012

### lingualatina

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.

3. Nov 11, 2012

### Abhinav R

T stands for Tension which is given by the downward component mg

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