3 Blocks Create a Tension Force- Find Acceleration

In summary: Right. That makes sense. I suppose I'm thinking horizontally (a different problem I had to do). Thank you! Now, is it possible to include the third equation? Or should I just do the two separately. My fear is that the accelerations won't equal. I'll try it out right now.
  • #1
Phoenixtears
83
0

Homework Statement


The coefficient of kinetic friction between the m = 2.6 kg block in Figure P8.35 and the table is 0.28. What is the acceleration of the 2.6 kg block? (The image is attached)
______m/s2


Homework Equations



Fk= (Mu)(N)
2nd Law statements

The Attempt at a Solution


I began by drawing three force diagrams, one for each block. Using the 2.6 and 3 blocks I wrote out 2nd law statements (because the block is shifting right), and substituted in variables for the friction forece equation:

Ca= T- Cg
Ba= T- Fk
Fk=(Mu)(Bg)... (According to my force diagrams , normal force equals weight force)

Then I sovled for T on the first equation. Substituting that in for T in the second, and replacing the Fk with the other equation:

Ba= Ca+ Cg- (Mu)(Bg)

I then solved for a. After plugging in all the numbers, I got a very high number as my a. Near 55, if I recall. What exactly did I do wrong?
 

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  • #2
Phoenixtears said:
Then I sovled for T on the first equation. Substituting that in for T in the second,
The tension is different for each rope. (If not, there would be no net force on the middle block.)
 
  • #3
Doc Al said:
The tension is different for each rope. (If not, there would be no net force on the middle block.)

But, according to the 3rd law, the tension between the 3 block and the 2.6 block would be equal. So why would the substitution not work?

Thanks in advance!
 
  • #4
Phoenixtears said:
But, according to the 3rd law, the tension between the 3 block and the 2.6 block would be equal.
Right, I misinterpreted what T you were talking about.
So why would the substitution not work?
Because there are two tension forces on the middle block.

Phoenixtears said:
Ba= T- Fk
You left out one of the tension forces.
 
  • #5
Doc Al said:
Because there are two tension forces on the middle block.


You left out one of the tension forces.

Right. That makes sense. I suppose I'm thinking horizontally (a different problem I had to do). Thank you! Now, is it possible to include the third equation? Or should I just do the two separately. My fear is that the accelerations won't equal. I'll try it out right now.
 

1. How do you define tension force?

Tension force is a pulling force that occurs when an object is being pulled from either end. It is a reaction force that is created by a string, rope, or cable when it is pulled tight.

2. What are the factors that affect the tension force in a system of 3 blocks?

The tension force in a system of 3 blocks is affected by the weight of the blocks, the angle at which the blocks are pulled, and the friction between the blocks and the surface they are resting on.

3. How does the number of blocks in a system affect the tension force?

The number of blocks in a system does not affect the tension force, as long as the blocks are connected by a single string or rope. The tension force remains the same throughout the system.

4. How can the acceleration of the system be calculated from the tension force?

The acceleration of the system can be calculated by dividing the tension force by the total mass of the system. This can be represented by the equation a = F/m, where a is acceleration, F is tension force, and m is the mass of the system.

5. What are some real-life applications of a system of 3 blocks creating a tension force?

A system of 3 blocks creating a tension force can be seen in various scenarios, such as a pulley system, a crane lifting heavy objects, or a person pulling a sled. It is also used in engineering and construction, such as in bridges and suspension systems.

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