Simple Pulley Problem: Finding Tension and Acceleration for Two Connected Blocks

[travikk]

Homework Statement

Two blocks with masses m = 2kg and M = 3kg are connected by a cord that passes over a
massless and frictionless pulley. The system is released from rest. Find the tension T in the cord and acceleration of the masses.

Homework Equations

F = m*a
Q = m*g

The Attempt at a Solution

I actually made a sketch of a system but I am not sure, If equations are correct. Could you check and show me the correct way to handle this simple problem? Thanks.

 

[Stonebridge]

If it's a light string and a frictionless pulley, then T1 = T2 = T
The tension is the same throughout the string.
You need to apply Resultant force = mass x acceleration for the two masses separately.
Both masses also have the same acceleration magnitude (=a) as the string is not extensible. (I guess!)
For the left mass, the resultant force is T-Mg
The mass is M and the acceleration is a
Do the same for the right mass.

 

[travikk]

Thanks for your answer!
Is it this way then:

 

[Stonebridge]

There are 2 ways of looking at this. The simple (quick) way, and the longer, more rigorous way.

The quick way will work here as it is a simple problem.
In this case, you can say that the resultant force is Q₁ - Q₂(Q₁ is greater than Q₂ and the 3kg mass will accelerate downwards, and the 2kg upwards.)
This resultant force accelerates the combined mass M+m
So your equation becomes
Q₁-Q₂ = (M+m)a

The more rigorous method is to apply F=ma separately to the two masses, as I said in my first post. (F is resultant force)

On the left
Q₁ -T₁ = Ma
On the right
T₂-Q₂ = ma

If you combine these two equations to eliminate T, you will get the same result as in the first method.
This 2nd method always works. The 1st method only works in simple cases such as this.
I recommend always approaching such problems using the 2nd method.

 

[rwisz]

I would first start by identifying the key variables in this problem. We have two masses, m and M, a tension force in the cord, T, and an acceleration, a. From the given information, we can also assume that the system is in a state of rest before it is released.

To solve this problem, we can use Newton's second law, which states that the force acting on an object is equal to its mass times its acceleration (F=ma). We can also use the equation for gravitational force (Fg=mg) to find the weight of each object, where g is the acceleration due to gravity (9.8 m/s^2).

Since the system is released from rest, we can assume that the acceleration of both masses is the same. Therefore, we can set up an equation for the net force acting on each mass:

For m: Fnet = T - Fg = ma
For M: Fnet = Fg - T = Ma

We also know that the masses are connected by a cord, so the tension force is the same for both masses. This allows us to set the two equations equal to each other:

T - mg = ma = Fg - T = Ma

Solving for T, we get T = (m+M)g/2 = (2+3)(9.8)/2 = 24.5N. This is the tension force in the cord.

To find the acceleration, we can plug in the value for T into either of the original equations and solve for a. Let's use the equation for m:

T - mg = ma
24.5 - (2)(9.8) = a(2)
a = 2.55 m/s^2

Therefore, the acceleration of both masses is 2.55 m/s^2.