What is causing the tension to decrease?

[greenskyy]

This is not a homework question, but simply a custom problem from class. I've attempted to answer it but realized I was incorrect, and can't think of an alternative answer.

Homework Statement

A cart is sitting on a frictionless track. A string is attached to this cart, which runs over a pulley with a mass tied on the other end. A hand is holding the cart in place.

When the cart is let go, and the mass allowed to fall, the force of the string on the cart slightly decreases. What is causing this to happen?

Homework Equations

f=ma

The Attempt at a Solution

Initially, I had thought that the force of the hand on the cart was causing additional tension in the string, but this could not be true. All the hand was doing was creating an equal and opposite force on the cart, causing it to remain at rest. The tension in the string should not have been affected by the hand. However, I can't think of any other explanation as to why this would occur. Any ideas / hints?

 

[RoyalCat]

Draw free body diagrams for the cart and mass in both situations.

One where the cart is held in place by the hand, and the other when the cart is allowed to move. Use Newton's Second Law to solve for the tension in the wire.

Things should become pretty clear once you've done that.

 

[greenskyy]

Oh my god, I never thought to draw a FBD for the mass as well. Let me answer this just in case anyone else is curious.

Because the mass is falling, we can reason that the net vertical force on the hanging mass is pointed down. Because weight is constant, the tension must have decreased.

Is this correct?

 

[RoyalCat]

Oh my god, I never thought to draw a FBD for the mass as well. Let me answer this just in case anyone else is curious.

Because the mass is falling, we can reason that the net vertical force on the hanging mass is pointed down. Because weight is constant, the tension must have decreased.

Is this correct?

Mathematically, you'd see that for the hanging mass:

\Sigma \vec F = m\vec g + \vec T

For the static situation, you'd get that the tension is equal to the weight of the mass. That makes sense, obviously.

Now let's assume we know the acceleration of the mass (You can calculate it, but that isn't necessary for the sake of this exercise, though it is a good exercise in its own right)
Let's call it \vec A, obviously it's pointed in the same direction as m\vec g

Let's look again at Newton's Second Law for the hanging mass:

m\vec A=m\vec g +\vec T

\vec T = m(\vec A-\vec g)

In scalar form:
T=m(g-A)

Now that you have this result, what can you say about what happens once the mass is released?

Think about how the force is distributed. In the static case, it all goes into the tension because the cart is unable to move, but in the dynamic case, what is the force of the weight of the mass responsible for? How is it distributed compared to the static case?

 

[greenskyy]

Once the cart is released, there is acceleration. Plugging it into that equation would cause T to be smaller than if A was 0.

The force of the weight of the mass is responsible for downward acceleration. How this is "distributed" does not make sense to me. It is a constant force regardless of the situation, is it not?

 

[RoyalCat]

Once the cart is released, there is acceleration. Plugging it into that equation would cause T to be smaller than if A was 0.

The force of the weight of the mass is responsible for downward acceleration. How this is "distributed" does not make sense to me. It is a constant force regardless of the situation, is it not?

Intuitively, the way I think about it is that in the static situation, the force is balanced by the tension in the string alone, while in the dynamic situation, part of the force goes to accelerating the cart, while a part of it is still balanced by the tension.

Well, mathematically you've got it right. From here on out it's just a question of settling it with your natural intuitions. Maybe someone more eloquent than I am can chime in on the subject.