Calculating Force F Required for Frictionless Blocks to Remain at Rest

[annatar]

Homework Statement

What is the magnitude of force F (which is pushing from the left on M) required to keep M1 and M2 relatively at rest to M?

Ignore all friction.

See attached diagram

Homework Equations

Newton's laws

The Attempt at a Solution

Well...If M is at rest, M1 and M2 will have an acceleration due to M2's weight.
When force F appears though, M2 will be pushed horizontally by M since they are in contact.

That's all I know atm.

 

[Astronuc]

The question asks about a force on M that allows "M1 and M2 to remain relatively at rest to M".

So what is the condition between M1 and M, such that M1 stays at rest with respect to M? If M1 does not move with respect to M, the M2 doesn't fall.

 

[nlightner]

To calculate the required force F, we can use Newton's second law: F = ma, where F is the force, m is the mass, and a is the acceleration. Since we are assuming a frictionless system, we can ignore any forces due to friction and focus on the forces acting in the horizontal direction.

First, we need to determine the acceleration of the system. Since M1 and M2 are at rest, we know that their acceleration is zero. However, M2 will experience an acceleration due to M's weight, which we can calculate using Newton's second law: a = Fg/M2, where Fg is the force of gravity and M2 is the mass of M2.

Next, we need to consider the forces acting on M1 and M2 in the horizontal direction. M1 is only being pushed by the force F, so its acceleration is equal to F/M1. M2, on the other hand, is being pushed by F as well as being pulled by M's weight. Therefore, its acceleration is equal to (F - Fg)/M2.

Since we want M1 and M2 to remain at rest, their accelerations must be equal. This means that the two equations we just calculated must be equal, so we can set them equal to each other and solve for F:

F/M1 = (F - Fg)/M2

Simplifying this equation, we get:

F = Fg(M1 + M2)/M2

Plugging in the values for Fg, M1, and M2, we can calculate the required force F:

F = (9.8 m/s^2)(3 kg + 5 kg)/5 kg = 11.76 N

Therefore, a force of 11.76 N is required to keep M1 and M2 at rest relative to M in a frictionless system.