Solving a 2-Mass System Connected by a Rope

[raptik]

Homework Statement

Two masses M₁ = .600kg and M₂ = 2.2kg connected by a cord of negligible mass and passes over a frictionless pulley of negligible mass. Assuming that y-axis has a positive upward direction, what is the tension in the cord?

Homework Equations

F = ma

W = mg

The Attempt at a Solution

I first tried to add up the F_g for both masses under the assumption that the opposing force would be the tension.

I then subtracted the larger F_g from the smaller F_g which was also wrong.

I then came upon https://www.physicsforums.com/showthread.php?t=201258" and tried to follow the process discussed there. I tried to solve for a based on the posts discussed in the other thread but was lost on posts 9 and 10 (NEwayz, back to what I did).

I found the equations of the two blocks to be T - m₁g = m₁a and m₂g - T = m₂a.
this makes T = m₁a + m₁g = m₂g - m₂a.
Plugging in the numbers gives 5.88N + .6a = 21.56N - 2.2a
solving for a gives a = 5.6ms^-2
Plugging this back into the T equation gives a T of 9.24N
I then double the T because there is T working on both ends of the rope and get 18.48N.
I turns out that this is an incorrect answer and I am now out of ideas. Please help.

 

[hage567]

You don't double T. The tension T is acting throughout the rope. It isn't T on one side and T on the other giving 2T total.

 

[thomate1]

I would approach this problem by first identifying all the given information and variables, and then using the relevant equations to solve for the unknown variable, which in this case is the tension in the rope.

Firstly, I would draw a free body diagram for each mass, showing all the forces acting on them. For M1, the forces would be the tension in the rope (T) and the force of gravity (mg), both pointing downwards. For M2, the forces would be the tension in the rope (T) and the force of gravity (mg), both pointing upwards.

Next, I would use Newton's second law, F=ma, to set up equations for each mass. For M1, the equation would be T - mg = ma, and for M2, it would be mg - T = ma. Since the pulley is frictionless and has negligible mass, we can assume that the acceleration of both masses is the same.

Then, I would solve for the tension in the rope by setting the two equations equal to each other and solving for T. This would give me T = (m1-m2)g/2.2. Plugging in the given values, we get T = (0.600-2.2)(9.8) / 2.2 = -10.56N. Note that the negative sign indicates that the tension in the rope is acting in the opposite direction to the assumed positive direction of the y-axis. This makes sense since the rope is pulling upwards on M2 and downwards on M1.

Finally, I would double this value to account for the tension acting on both ends of the rope, giving a final answer of 21.12N for the tension in the rope.

In conclusion, as a scientist, I would approach this problem by carefully analyzing the given information, using relevant equations, and following a systematic and logical approach to arrive at the correct solution.