Solving for Tension in Two Block System

In summary, two blocks of equal mass are connected by a horizontal string and hang from a ceiling. The tension in the horizontal string, T1, can be found by using the equations ΣFx and ΣFy. By setting ΣFy to equal mass times acceleration, Ta and Tb can be found and used to solve for T1 in the equation TbT1 / (2mg-Tb). However, this expression for the x direction is incorrect and instead a specific point should be chosen to write the horizontal equation.
  • #1
Inertialforce
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Homework Statement


Two blocks of equal mass hang vertically from a ceiling as shown in the diagram. A horizontal string connects the two vertical strings. Determine the tension T1 in the horizontal string. All strings are considered to be massless.


Homework Equations


ΣFx and ΣFy


The Attempt at a Solution


m1 = m2 (therefore I referred to the masses as just "m")
angle = 90(degrees)

ΣFy = may
Ta + Tb -mg -mg = 0
Ta + Tb - 2mg = 0
Ta + Tb = 2mg
Ta = 2mg - Tb


ΣFx = max
TaT1 - TbT1 = 0
TaT1 = TbT1
(2mg - Tb)T1 = TbT1
T1 = TbT1 / (2mg-Tb)

I got this question wrong and since there are no numerical values given I am unsure as to where I am going wrong.
 

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  • #2
Inertialforce said:

Homework Statement


Two blocks of equal mass hang vertically from a ceiling as shown in the diagram. A horizontal string connects the two vertical strings. Determine the tension T1 in the horizontal string. All strings are considered to be massless.


Homework Equations


ΣFx and ΣFy


The Attempt at a Solution


m1 = m2 (therefore I referred to the masses as just "m")
angle = 90(degrees)

ΣFy = may
Ta + Tb -mg -mg = 0
Ta + Tb - 2mg = 0
Ta + Tb = 2mg
Ta = 2mg - Tb

I would suggest you also draw the freebody diagram for just one mass; then you can find the tension in each string Ta and Tb (though I don't think you need them for this problem).

ΣFx = max
TaT1 - TbT1 = 0
TaT1 = TbT1
(2mg - Tb)T1 = TbT1
T1 = TbT1 / (2mg-Tb)

This expression for the x direction does not look right. For example, you can see that you can just cancel out the T1 completely from any of these equations. Also, the expression TaT1 - TbT1 = 0 does not have the right units to be a force equation, and Ta and Tb do not have any components in the horizontal direction.

Instead, when you are writing the horizontal equation, pick a specific point (for example, one knot where strings come together) and write the horizontal equation for that. What do you get?
 
  • #3
Can you provide any insight?

I would first like to commend you on your attempt at solving this problem using the correct equations and principles. However, there seem to be some errors in your calculations.

Firstly, when you set up your ΣFy equation, you have correctly identified that the sum of the vertical forces must equal the mass times acceleration in the y-direction. However, the equation should be Tb + Tb - 2mg = 0, as both blocks are experiencing the same tension force in the vertical strings. This should then lead to Tb = mg.

Next, when setting up your ΣFx equation, you have correctly identified that the sum of the horizontal forces must equal zero. However, the equation should be Ta - Tb = 0, as the tension in the horizontal string is equal to both Ta and Tb, and they must balance each other out. This should then lead to Ta = Tb.

Finally, when solving for T1, you have correctly set up the equation TaT1 - TbT1 = 0. However, when substituting in values, it should be (mg)(T1) - (mg)(T1) = 0, as Ta and Tb are both equal to mg. This would then lead to T1 = 0, which is the correct answer.

In conclusion, it seems that the main error in your calculations was assuming that Ta and Tb were equal to 2mg, when in fact they are equal to mg. This led to incorrect values and ultimately, an incorrect answer. I hope this helps clarify any confusion and helps you approach similar problems in the future.
 

FAQ: Solving for Tension in Two Block System

What is tension in a two block system?

Tension in a two block system refers to the force exerted on each block by the connecting rope or string. It is the force that keeps the blocks together and prevents them from separating.

How is tension calculated in a two block system?

Tension can be calculated by using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In a two block system, the net force is equal to the tension force, so by rearranging the equation, we can solve for the tension.

What factors affect the tension in a two block system?

The tension in a two block system is affected by the mass of the blocks, the angle at which the rope is pulled, and the acceleration of the blocks. A larger mass or a steeper angle will result in a higher tension, while a smaller acceleration will result in a lower tension.

How does the direction of the tension force change in a two block system?

In a two block system, the direction of the tension force is always in the direction opposite to the motion of the blocks. This is because the tension force is pulling the blocks towards each other and preventing them from separating.

Can the tension in a two block system ever be zero?

No, the tension in a two block system can never be zero as long as the blocks are connected by a rope or string. This is because even if the blocks are at rest, the tension force is still present to keep the blocks together and maintain their position.

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