What is the flaw in my logic for solving this Newton's 2nd law problem?

[David112234]

Homework Statement

http://postimg.org/image/z9uzlyt3p/

Homework Equations

F=MA
W=MG

The Attempt at a Solution

the force on the block must be gravity *its mass (weight) since it hangs, but because of the pulley and other box pulling the other way, creating an T, the force on the hanging block should be -MG - T (from M2).
-Mg because the wight is hanging down
I tried T-MG, but it was wrong as well

It tells me it is wrong, what is the flaw in my logic?

 

[Ken G]

I think I understand the situation-- it is in their strange x and y coordinates. Apparently they intend x and y to be the same thing-- advancement of the system such that block 2 goes up and block 1 goes down. It's not what they say, so it's a poor problem, but it would make sense as a way of expressing the main constraint on the system. This means down is the positive direction, so the answer is mg-T. Then the acceleration A is the same for both blocks, which is presumably how they want their constraint to work out.

 

[David112234]

I originally tough the same, but it is also not the correct answer

 

[Ken G]

Well, what we can say for sure is T and mg are in opposite directions, and they are the only forces on block 1, so you can be sure it is either T-mg or mg-T. It all depends on what sign convention their "y" coordinate is using, but since they don't say, you have to guess. The sensible one is the same as x, so that would make mg-T correct. It sounds like mg+T is the only one you haven't tried, but there's no universe where that answer makes any sense at all, unless they are imagining that the sign of the number that substitutes for T is positive at the left end of the rope and negative at the right end, which would be a pretty unusual way to look at the situation!

 

[SammyS]

the force on the block must be gravity *its mass (weight) since it hangs, but because of the pulley and other bo

Homework Statement

http://postimg.org/image/z9uzlyt3p/

Homework Equations

F=MA
W=MG

The Attempt at a Solution

the force on the block must be gravity *its mass (weight) since it hangs, but because of the pulley and other box pulling the other way, creating an T, the force on the hanging block should be -MG - T (from M2).
-Mg because the wight is hanging down
I tried T-MG, but it was wrong as well

It tells me it is wrong, what is the flaw in my logic?

It clearly states that up is to be considered as positive for part (E).

Is the rope pulling up on Block 1 or is it pushing Block 1 down?

It should be T - m₁g .

Are you sure you didn't have a typo when you tried that.

 

[David112234]

. It sounds like mg+T is the only one you haven't tried, but there's no universe where that answer makes any sense at all, unless they are imagining that the sign of the number that substitutes for T is positive at the left end of the rope and negative at the right end, which would be a pretty unusual way to look at the situation!

I got it, it turns out its -mg+T . The reasoning behind it is because even tho the mass on the left is pulling town, the tension is redirected as up, it pulls the hanging mass up, and gravity down. Thanks for the help.

 

[Ken G]

Yes, it does make sense since they did specifically say that the +y direction is up, as Sammy5 pointed out. I thought that couldn't be the case because you said you had tried that one, but you must have made an oversight. Personally, I would not have used that +y convention, I would have made the +y direction be down, since then a_1 = a_2 is the key constraint in solving for the motion. They would have to use a_1 = - a_2 when they solve it, which I find unnecessarily awkward, but certainly not wrong.