Can't get this tension derivation........

In summary, the conversation discusses the concept of tension in relation to an Atwood machine and two unequal masses attached to a rope. It is noted that the tension in the rope is not always equal to the weight being suspended, and that it can vary depending on the direction and acceleration of the masses. The concept is further explained through the use of free body diagrams and the calculation of net forces.
  • #1
singh_abhi
2
0
Ok guys, i just don't seem to know anything about tension, i thought that it would always equal the weight that is suspended from the rope...But that does not seem to be the case...and also i don't get the issue of its direction.
Anyways, suppose we have an atwood machine and two unequal masses attached to a rope...then the system accelerates...What I don't get is the fact that here Tension is not equal to any of the weight, why is it not so...When we suspend a single weight from a ceiling, then it is...i think the reason i am confused is that i have no understanding of the fundamentals of tension...Any help will be appreciated. Thanks...
 
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  • #2
singh_abhi said:
Ok guys, i just don't seem to know anything about tension, i thought that it would always equal the weight that is suspended from the rope...But that does not seem to be the case...and also i don't get the issue of its direction.
Anyways, suppose we have an atwood machine and two unequal masses attached to a rope...then the system accelerates...What I don't get is the fact that here Tension is not equal to any of the weight, why is it not so...When we suspend a single weight from a ceiling, then it is...i think the reason i am confused is that i have no understanding of the fundamentals of tension...Any help will be appreciated. Thanks...
Are you familiar with drawing free body diagrams (FBDs)? They will show all of the forces on an object. If the rope is not just supporting the weight but is also providing acceleration upward, the tension force will be greater than the weight being accelerated...
 
  • #3
berkeman said:
Are you familiar with drawing free body diagrams (FBDs)? They will show all of the forces on an object. If the rope is not just supporting the weight but is also providing acceleration upward, the tension force will be greater than the weight being
accelerated...
I understand it, but only from a mathematical point of view, since there is an upward acceleration on the smaller block, and since weight acts downwards ,then the upward acceleration must be due to tension. I also know that the net force acting on the system would be m2g-m1g...and that this would equal (m2+m1)a...
I get why this happens conceptually...I just can't get why tension is less than the heavier weight and greater than the smaller weight.
 
  • #4
singh_abhi said:
I just can't get why tension is less than the heavier weight and greater than the smaller weight.

Look at the forces on each mass on it's own...

The heavier weight M1 is accelerating downwards suggesting that the tension in the rope isn't sufficient to support it's weight... T1 = M1(g-a)

The lighter weight M2 is accelerating upwards so the tension must provide an upward force greater than it's weight... T2 = M2(g+a)

If the same rope supports both masses (and any pulley is frictionless with negligible inertia) then the tension is the same in all parts of the rope so equate..

T1=T2
and
M1(g-a) = M2(g+a)
expand
M1g-M1a = M2g+M2a
rearrange
M1g-M2g = M1a+M2a
or ...
a = (M1-M2)g/(M1+M2)

Note that a is less than g which is what you would expect. The small mass stops the larger one falling at g.
 

Related to Can't get this tension derivation........

1. What is tension derivation in science?

Tension derivation is a process in which the force acting on an object is determined by analyzing the tension in a system of connected objects. This is commonly used in physics and engineering to calculate the forces involved in systems such as pulleys, ropes, or cables.

2. How is tension derivation calculated?

To calculate tension derivation, you must first identify all the forces acting on the system. Then, using Newton's Second Law (F=ma), you can calculate the net force acting on the system. Finally, using the knowledge that tension is a result of this net force, you can solve for the tension in the system.

3. What are some real-world applications of tension derivation?

Tension derivation has many practical applications in the real world. It is used in the design of bridges and other structures to determine the amount of tension that can be safely supported. It is also used in the study of human movement, such as analyzing the tension in muscles and tendons during physical activity.

4. Are there any limitations to tension derivation?

While tension derivation is a useful tool in physics and engineering, it does have its limitations. It assumes that all objects in the system are in a state of equilibrium, and it does not take into account factors such as friction or air resistance. Additionally, it may not be accurate for systems with rapidly changing forces or high speeds.

5. How can I improve my understanding of tension derivation?

To improve your understanding of tension derivation, it is important to have a strong grasp of Newton's Laws of Motion and basic mathematical principles. Practice solving problems and working with different types of systems to become more familiar with the process. Additionally, seeking out additional resources such as textbooks, online tutorials, and practice exercises can also be helpful.

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