How Much Force to Apply on m1 to Make m2 Jump Off the Table?

In summary, the conversation discusses the conditions for m2 to jump and leave the table, assuming no mass for the spring and no friction. It also mentions the role of Newton and Hooke's laws in solving this problem. The possible equations for determining the force needed for m2 to jump are also mentioned. The person is confused about the force needed to push down on m1, with a guess that it could be equal to the total force pulling on m2 or the force needed to support m1. The person also asks for clarification on the net force acting on m2 at the moment it jumps.
  • #1
thoff430
4
0

Homework Statement



Hey there,

We assume that the spring got no mass and there are no frictions. If you want to push down just as hard on m1 that if you release ... m2 will be just about to jump and leave the table?


unbenannt_979.jpg


Homework Equations



Newton and Hooke are our very best friends :)

The Attempt at a Solution



1. F > (m1 + m2) * g

or

2. F > m2 * g ?

Im a bit confused, as m1 is already pushing on the string even if there is no extra force... but how much do I have to push down for myself on m1? My guess is that 1 would be to total amount of force "pulling" on m2, but the force needed to get this right, would be 2, as m1 is supporting.

Thanks in advance
 
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  • #2
What is the condition that m2 jumps and leaves the table? What can you say about the net force acting on it at that instant?

ehild
 

FAQ: How Much Force to Apply on m1 to Make m2 Jump Off the Table?

What is the concept of "masses coupled with a spring"?

"Masses coupled with a spring" refers to a system in which two or more masses are connected by a spring, creating a physical coupling between them. This system is commonly used in physics to study the behavior of oscillating systems.

How does the mass affect the behavior of the system?

The mass of the objects connected by the spring determines the system's natural frequency of oscillation. A higher mass will result in a lower natural frequency, meaning the system will oscillate at a slower rate.

What is the role of the spring in this system?

The spring provides a restorative force to the system, causing the masses to oscillate back and forth. The stiffness of the spring (measured by its spring constant) determines how strong this restorative force is.

How does the amplitude of oscillation change with increasing mass?

As the mass of the objects increases, the amplitude of oscillation decreases. This means that the distance the masses move from their equilibrium position becomes smaller as the mass increases.

What factors can affect the behavior of this system?

Aside from the mass and stiffness of the spring, factors such as external forces, friction, and damping (the dissipation of energy) can also affect the behavior of the system. These factors can alter the natural frequency and amplitude of oscillation of the system.

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