Solving the Tension Equation: Understanding the Progression

[msudawgs267]

I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
v= [square root of] (Ft-mg)r /m

The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.


http://img3.imageshack.us/img3/4830/equationi.jpg


I hope my question was not written too unclear and any help would be amazing. Thanks!

 

[alphysicist]

Hi msudawgs267,

I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
v= [square root of] (Ft-mg)r /m

The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.

I would not say the expression on the left is progression from one step to another. The starting point is Newton's law in the form:

$$\sum F = m a$$

and then they are saying that

$$\sum F \to F_T - mg$$

and

$$m a \to m\frac{v^2}{r} \ \ \ \mbox{(for centripetal acceleration)}$$

Setting these two new expressions equal to each other and solving for v gives the answer.

 

[nlightner]

Thank you for your question about the tension equation and its progression. The equation you are referring to is known as the tension equation, which is used to calculate the tension in a string or rope. The equation can be written as T = Ft - mg, where T is the tension, Ft is the force applied to the string, m is the mass of the object attached to the string, and g is the acceleration due to gravity. This equation is derived from Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma).

To understand the progression from this equation to the one you mentioned (v=√(Ft-mg)r/m), we need to look at the concept of work and energy. Work is defined as the product of a force and the distance over which that force acts. In the case of a string, when a force is applied to one end, it will cause the other end to move a certain distance. This work done on the string is equal to the change in the kinetic energy of the object attached to the string.

Using this concept, we can derive the equation v = √(Ft-mg)r/m. The left side of the equation represents the final velocity (v) of the object, which is equal to the initial velocity (0) plus the change in velocity (√(Ft-mg)r/m). The right side of the equation represents the work done on the object, which is equal to the force applied (Ft-mg) multiplied by the distance over which it acts (r), divided by the mass of the object (m).

Therefore, we can see that the two equations are equivalent, as they both represent the same concept of work and energy. The first equation (T = Ft - mg) gives us the tension in the string, while the second equation (v = √(Ft-mg)r/m) gives us the velocity of the object.

I hope this explanation helps you understand the progression from one equation to the other. If you have any further questions, please do not hesitate to ask.