Two Blocks and a pulley (Equation derivation)

[Jayy962]

Homework Statement

Look at the figure below. Derive the formula for the magnitude of the force F exerted on the large block (mC) in the figure such that the mass mA does not move relative to mC. Ignore all friction. Assume mB does not make contact with mC.

Homework Equations

F = ma

The Attempt at a Solution

So mB is pulling the block mA to the right. Therefore if the block mC were to not move, mA would move to the right. So the acceleration of mA needs to be equal to the acceleration of mC to make sure the blocks stay at the same relative position. mA is accelerating at the rate of g since mB is accelerating at the rate of g.

To accelerate mC at the rate of g, you'd need to move both the mass of mC and mB therefore I thought the force would be (mB + mC)g.

This is wrong and I'm not sure why.

 

[BvU]

The flaw is at "since mB is accelerating at the rate of g" : the force that accelerates is mB * g alright, but the mass that has to be accelerated is mA+mB .

However, it is much better to use the floor (or table, or what is it) as a reference frame and set up free body diagrams for each of the three masses. Note that the connection bar between the pulley and block C exercises a force on C as well.

 

[HallsofIvy]

The way the situation is shown in your picture, this is impossible. As mass mC accelerates downward, mass mB will have an acceleration relative to the pulley. But the pulley is attached to mass mC so any movement relative to the pulley is motion relative to mass mC. Accelerating mC also accelerates the pulley and so adds more acceleration to mA.

 

[BvU]

Well, have to get used to the new environment.
Ivy makes things complicated. Let's assume no swinging of block B.
Make the three drawings. All have the same acceleration a, as a resultant of different forces.
$\vec a$ is horizontal.
On block A:
On block B:
On block C:
Fill in the blanks!

 

[HalfLostAndSearching]

I can provide a more detailed explanation for the correct derivation of the formula for the magnitude of the force F exerted on the large block mC.

First, let's consider the free body diagram of the system:

mB is pulling on mA with a force T, and the weight of mC is acting downwards with a force mg. Since the system is in equilibrium, the net force on mC must be zero. This means that the force F must be equal in magnitude and opposite in direction to the weight of mC, so we can write:

F = -mg

Now, we also know from Newton's second law (F = ma) that the acceleration of an object is directly proportional to the net force acting on it. In this case, the net force on mC is the force F, so we can write:

F = maC

Combining these two equations, we get:

-mg = maC

Rearranging for F, we get the final formula:

F = maC = mCg

This is the correct formula for the magnitude of the force F exerted on mC in order to keep the system in equilibrium. It takes into account the weight of mC, which needs to be balanced by the force F in order to prevent it from moving. The weight of mB does not need to be considered since it is not directly involved in the equilibrium of the system.

I hope this explanation helps to clarify the correct derivation of the formula for F. Remember, as a scientist, it's important to carefully consider all the forces acting on a system and use the appropriate equations to accurately describe its behavior.