How Can You Calculate Gravity Using an Atwood Machine?

[TroubledStudent]

There is a problem in my physcs book that I have been trying to solve for some time now, and I just can't get trough it. If someone could help me a little bit, it would really be great.
My question is about the Atwood machine.

There are two blocks of same weight (M) suspended on a rope of each side of a pulley. There is a squared plaque (weight = m) placed on one of the blocks. When the block is dropped, is accelerates on a distance H until the squared plaque is stopped by a ring, but the block underneath continues it way at a constant speed. The distance D traveled by the block at constant speed, after the plaque is stopped by the ring, lasts "t" seconds. Prove that gravity acceleration, "g", is represented by the formula:

g = ((2M + m)D^2) / (2mHt^2)

Help for this question would be really, really apprecied...thank you !

Christopher

 

[HallsofIvy]

The sentence "The distance D traveled by the block at constant speed, after the plaque is stopped by the ring, lasts "t" seconds." cofused me. It really should be "it took t seconds for the block to travel distance D ..." since there will be nothing to stop the block.

The two M blocks balance so the total force, after m is placed on, is -mg. The total mass of the system is 2M+ m so the acceleration, from -mg= (2M+m)a is -mg/(2M+m).

The distance H move in time t₁ is: H= (-mg/(2(2M+m))t₁² ((1/2)a t²) and the speed at time t is V= (-mg/(2M+m))t₁.

Solve the first equation for t₁ and plug that into the equation for V so you know the speed after m is removed.

Once you have V in terms of g, m, M, use

D= Vt and solve for g.

 

[M98Ranger]

Sure, I would be happy to help you with this problem. The Atwood machine is a classic physics problem that involves the principles of mechanics and Newton's laws of motion. Let's break down the problem and go through the steps to solve it.

First, we need to understand the setup of the Atwood machine. As you described, there are two blocks of equal weight (M) suspended on either side of a pulley. The pulley is frictionless, meaning there is no resistance as it rotates. On one of the blocks, there is a squared plaque with a weight of m. When the block is dropped, it accelerates a distance H until the squared plaque is stopped by a ring. The block underneath continues to move at a constant speed, covering a distance D in time t.

Now, let's consider the forces acting on the system. The two blocks are connected by a rope, so they experience equal and opposite tension forces. The block with the squared plaque also experiences an additional force due to the weight of the plaque. This force is given by F = mg, where g is the acceleration due to gravity. The block without the plaque only experiences the tension force.

Next, we can apply Newton's second law, which states that the net force on an object is equal to its mass times its acceleration (F = ma). In this case, we have two objects with different masses, but they are connected and move together, so we can treat them as one system. This means that the net force on the system is equal to the sum of the individual forces acting on the blocks.

Since the blocks are connected by a rope, the tension forces cancel out. This leaves us with the force due to the weight of the plaque (mg) and the force due to the acceleration of the system (ma). We can set these equal to each other and solve for the acceleration (a).

mg = ma

a = g

This tells us that the acceleration of the system is equal to the acceleration due to gravity (g). Now, let's consider the distance traveled by the block with the plaque. We can use the equation for constant acceleration (x = xo + vot + 1/2at^2) to find the distance traveled.

x = 0 + 0t + 1/2gt^2

x = 1/2gt^2

Since this distance is equal to H, we can substitute H for x and solve for