Question

While reading "Physics for Engineers and Scientists", I came upon this example which I worked through and then ended up with a question which is attached:

Answer 1

Start by fixing the second equation you wrote down by choosing a coordinate system. I suggest +y to be downward. Then you get ma = -T + mg. Alternatively, with +y pointed upward, -ma = T-mg. Go from there on, making sure the direction of your acceleration makes sense...

Answer 2

The diagrams are from the book so are the equations. We are told that the tension in a cable/string with no mass and a frictionless pulley is the same at all points; however using the textbook's numbers to get the m2 to move at an acceleration of .001g I would need a tension of .26 m1; HOWEVER to get that same acceleration on m1, I would need a tension of .25 m1. The "tensions" would be pointing in different directions however their magnitude would be different. That is my confusion.

Answer 3

And to add my third comment to my own question; it seems to me that the book is leaving out of the free body diagram of m2 the acceleration of the whole system which is given as .001 m/sec^2.

Answer 4

The first equation is wrong. The correct equation should be:
\[m_1 a_{1x} = m_1 g \sin(\Theta) - T \cos(\Theta)\]

The equation is concerned with the x-direction of the accelerations and forces, hence we should only talk about the x-direction of the force T as well.

Answer 5

I was DUMB!! I forgot forces are vector even after happytreefriend7 pointed it out to me; so I dumped both the Tension and m2g in the second equation ignoring that they were pulling in different directions - one needed to be negative.