Changing masses on an atwood machine

In summary, the conversation discusses a problem involving a frictionless, massless pulley holding two equal masses and an additional mass being added to one side. Initially, the system is in equilibrium with a net force of 0 and no acceleration. However, after the additional mass is added, the system is no longer in equilibrium and mass 1 accelerates downwards at approximately 1.8 m/s^2. The tension in the string is also calculated to be 58N. The conversation also raises questions about the system's constant acceleration after equilibrium is disturbed and how the situation may differ if observed from two different events.
  • #1
overthewater
1
0
An example of this kind of a problem would be a frictionless, massless pulley holding two equal masses(say, 5kg each). But then, an additional mass is added to one side (ie, a frog jumping on to mass 1, 1 kg). What would happen to the system? What if the frog jumps off?Initially, I know that the system has a net force of 0. F(net) = T -Fg while T=Fg.
It is not accelerating.
But after the frog jumps on, the system is no longer in equilibrium and mass 1 accelerates downward because the force of gravity on the "new" mass 1 (5+1= 6 kg) is larger than the tension in the string . So, mass 1 would be accelerating downwards at approximately 1.8 m/s^2. This is because a = (m1*g-m2*g)/(m1+m2). And tension, calculated using mass 2 where T>Fg would be T=Fnet+Fg=5*1.8+5*9.8= 58N.


But I don't think that the actual test question would be this simple. Is there anything different I need to look at? For example, does the system still have a constant acceleration just after the equilibrium is disturbed? Does the fact that a mass was added during our observations cause the situation to differ from if we were to look at it from two different events (ie, before the frog, when the masses were equal, and then a situation where one mass is greater than the other. Then comparing the two events)
 
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  • #2
Hello water, welcome to PF :)
Use the template: it's mandatory in PF and it helps you to order your thinking.

Check your calculation: g/11 is not 1.8 m/s2.
(Otherwise, what you do seems correct to me).

Read what you typed before posting; your last paragraph is totally unclear.

And when the frog jumps off (sideways): what is then the net force on each mass ?
 

What is an Atwood machine?

An Atwood machine is a simple device used to demonstrate the principles of classical mechanics. It consists of two masses connected by a string or rope that passes over a pulley. The two masses are usually unequal, and the system is designed to show how the difference in mass affects the acceleration of the system.

How do you change the masses on an Atwood machine?

To change the masses on an Atwood machine, you can simply replace the existing masses with new ones of different weights. Make sure to securely attach the masses to the string or rope and ensure that the pulley is functioning properly.

Why would you want to change the masses on an Atwood machine?

Changing the masses on an Atwood machine allows you to observe and study how different masses affect the acceleration of the system. This helps to understand the principles of classical mechanics, such as Newton's laws of motion and the concept of force.

What factors affect the acceleration of an Atwood machine?

The acceleration of an Atwood machine is affected by several factors, including the difference in mass between the two masses, the force of gravity, and the friction between the string/rope and the pulley. Other external factors, such as air resistance, may also affect the acceleration of the system.

How do you calculate the acceleration of an Atwood machine?

The acceleration of an Atwood machine can be calculated using the equation a = (m1 - m2)g / (m1 + m2), where m1 and m2 are the masses of the two objects and g is the acceleration due to gravity. This equation assumes that the pulley is massless and there is no friction or air resistance. In real-world situations, these factors may need to be taken into account for a more accurate calculation.

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