Question

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two blocks of masses m1 = 1.5 kg, and m2 = 2.4 kg are connected by a massless string which passes over a frictionless pulley of negligible mass, as shown above.
What is the acceleration of the system? m/s2

What is the magnitude of the tension in the string? N

Answer 1

Answers:

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

$F=ma$   (1)

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$.

We will also need to draw the Free Body Diagram of each block in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (See figure attached).  

We already know $m_{2}$ is greater than $m_{1}$, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$; therefore the acceleration of the system will be in the direction of $P_{2}$, as shown in the figure attached.

We also know by the information given in the problem that the pulley does not have friction and has negligible mass, and the string is massless.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the acceleration of gravity $g=9.8\frac{m}{s^{2}}$  acting on the blocks:

$P=mg$   (2)

For block 1:

$P_{1}=m_{1}g$   (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$    

$P_{1}=14.7N$   (4)

For block 2:

$P_{2}=m_{2}g$   (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$    

$P_{2}=23.52N$      (6)

Then, we are going to find the acceleration $a$ of the whole system:

$F_{r}=P_{1}+P_{2}$   (7)

$P_{1}+P_{2}=(m_{1}+m_{2})a$   (8)

Where the Resulting Force $F_{r}$  is equal to the sum of the weights $P_{1}$ and $P_{2}$.  

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$, this means it must have a negative sing; while $P_{2}$ is in the same direction of $a$.

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$  

$8.82N=(3.9kg)a$  

Then:

$a=\frac{8.82N }{3.9kg}$  

$a=2.26\frac{m}{ s^{2}}$  

This is the acceleration of the system.

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, the result will be the same.  

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the acceleration is in the same direction of the tension of the string, so:

$F_{r}=T-P_{1}$   (9)

$T-P_{1}=m_{1}a$   (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$    

Then;

$T=18.09N$   This is the tension of the string

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must have a negative sign, so:

$F_{r}=T-P_{2}$   (9)

$T-P_{2}=m_{2}a$   (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$    

Then;

$T=18.09N$    This is the same tension of the string