How Do You Calculate Tension in a System of Connected Blocks?

[tigerlili]

Homework Statement

three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude T3 = 13.8 N. If m1 = 15.1 kg, m2 = 24.8 kg, and m3 = 30.5 kg, calculate (a) the magnitude of the system's acceleration, (b) the tension T1, and (c) the tension T2.

(the figure has 3 blocks in this order: 1----2-----3--- being pulled by a hand on this side

Homework Equations

f=ma

The Attempt at a Solution

well, i got part a. that was just the sum of the masses, using t3 as the force in the f=ma equation. now, for the next part, i would think that the tension on block 2 would be the sum of block 1 and 2 (because they go in decreasing order) times the acceleration. likewise, i would think that block 1's tension would just be... the mass of block one times the acceleration. evidently that is wrong, so I'm just wondering if someone can give me a hand. thanks!EDIT:
..this was actually correct
just, the way i was entering it was wrong
ignore this post, sorry! :)

 

[anathema]

Thank you for your question. It seems that you have already solved part (a) correctly. For part (b), you are on the right track. The tension on block 1 would indeed be the mass of block 1 times the acceleration. However, for the tension on block 2, it would be the sum of the masses of blocks 1 and 2 times the acceleration, not just the mass of block 2. This is because both blocks are being pulled by the same force, so the tension on block 2 would include the weight of block 1 as well.

For part (c), you can use the same logic as part (b) to calculate the tension on block 3. It would be the sum of the masses of blocks 1, 2, and 3 times the acceleration.

I hope this helps. Let me know if you have any further questions. Keep up the good work!
Scientist

 

[tomcue]

Hi there,

I would like to provide a more detailed response to your question. Firstly, let's break down the problem and identify the key concepts involved.

1. The blocks are connected, which means they are all moving together as one system.
2. The force T3 is acting on the entire system, not just on block 3.
3. The table is frictionless, so there is no external force acting on the system.

Now, let's look at the equations that we can use to solve this problem.

1. Newton's Second Law: F = ma, where F is the net force acting on the system, m is the total mass of the system, and a is the acceleration of the system.
2. Newton's Third Law: For every action, there is an equal and opposite reaction. This means that the tension forces acting on each block will be equal and opposite to the force T3.

Using these equations, we can solve for the magnitude of the system's acceleration (a) in part (a) of the problem. We know that the net force acting on the system is T3, so we can plug that into the formula F = ma. This gives us:

T3 = ma

Solving for a, we get:

a = T3/m

a = 13.8 N / (15.1 kg + 24.8 kg + 30.5 kg)

a = 0.137 m/s^2

Now, let's move on to part (b) and (c), where we need to calculate the tensions T1 and T2. As mentioned earlier, the tension forces acting on each block will be equal and opposite to the force T3. So, we can use Newton's Third Law to solve for T1 and T2.

For block 1, we have:

T1 = m1 * a

T1 = 15.1 kg * 0.137 m/s^2

T1 = 2.07 N

For block 2, we have:

T2 = m2 * a

T2 = 24.8 kg * 0.137 m/s^2

T2 = 3.40 N

Therefore, the tensions T1 and T2 are 2.07 N and 3.40 N, respectively.

I hope this explanation helps to clarify any confusion you may have had. Remember to always clearly identify