3 masses, 1 pulley dynamics question

[karlmarxdumb]

Homework Statement

mass 1 sits on a table, mass 2 sits on top of mass 1, mass 3 is connected to mass 2 via a string and pulley. (the pulley extends diagonally from the edge of mass 1)
mass 3 is hanging beside mass1 and its side in full contact with mass 1.
with what force must you push mass 1, such that mass 3 doesn't fall?
all surfaces are frictionless.
m1 = 10kg m2=5kg m3=8kg

Homework Equations

force of contact between 1 and 3 are the same in magnitude
mass 1 moves because of the applied force, mass two because of the tension in the string, and mass three because it is contact with 1, and is being pushed by one.
relative to each other, the masses don't move.

The Attempt at a Solution

i'm not sure how this can be done. I've tried several ways, but i don't understand what could keep mass 3 from falling.


i really appreciate any help!

 

[anathema]

Thank you for your question. I can understand your confusion about how to keep mass 3 from falling in this scenario. Let me explain the physics behind it.

Firstly, let's consider the forces acting on each mass separately. Mass 1 has a force of gravity pulling it towards the ground, which we can represent as mg, where g is the acceleration due to gravity (9.8 m/s^2). Since the surface is frictionless, there is no force of friction acting on mass 1.

Mass 2 also has a force of gravity pulling it towards the ground, mg, but it also has a force of tension in the string pulling it towards mass 3. This tension force is equal in magnitude to the force of gravity, since the masses are not accelerating relative to each other. This means that the tension force is also mg.

Now, let's consider mass 3. It has a force of gravity pulling it towards the ground, mg, but it is also in contact with mass 1. This means that there is a normal force acting on mass 3 from mass 1, which is equal in magnitude to the force of gravity, mg.

Since we want to keep mass 3 from falling, we need to make sure that the normal force from mass 1 is equal to or greater than the force of gravity on mass 3. In other words, we need to push mass 1 with a force of at least mg.

However, since mass 3 is also being pulled by the tension force, we need to make sure that the tension force is also greater than the force of gravity on mass 3. This means that we need to push mass 1 with a force greater than mg, but less than 2mg (since the tension force is also mg).

So, to summarize, to keep mass 3 from falling, you need to push mass 1 with a force greater than mg, but less than 2mg. I hope this helps clarify the situation for you. Let me know if you have any further questions.

 

[baba89]

I would approach this problem by analyzing the forces acting on each mass and using the laws of motion to determine the necessary force to keep mass 3 from falling.

First, let's consider mass 1. Since it is on a table and all surfaces are frictionless, the only force acting on it is the force of contact from mass 2. This force has the same magnitude as the force of contact from mass 3, as stated in the problem. Therefore, we can say that the force acting on mass 1 is equal to the sum of the weights of masses 2 and 3 (mg2 + mg3).

Next, let's consider mass 2. It is being pushed by mass 1 and pulled by the tension in the string connected to mass 3. Since the string is frictionless, the tension force is equal to the weight of mass 3 (mg3). Therefore, the net force on mass 2 is the force of contact from mass 1 (mg1) minus the tension force (mg3).

Finally, let's consider mass 3. It is being pulled downwards by its weight (mg3) and pushed upwards by the force of contact from mass 1 (mg1). Since we want mass 3 to remain in contact with mass 1 and not fall, the force of contact from mass 1 must be equal to the weight of mass 3 (mg3).

Now, we can set up an equation to solve for the force needed to push mass 1:

mg1 = mg2 + mg3
mg1 = (5kg)(9.8m/s^2) + (8kg)(9.8m/s^2)
mg1 = 98N + 78.4N
mg1 = 176.4N

Therefore, in order to keep mass 3 from falling, we need to push mass 1 with a force of 176.4N. This will create a force of contact between mass 1 and 3 that is equal in magnitude to the weight of mass 3, keeping it in place.

I hope this explanation helps! Remember, as a scientist, it is important to carefully analyze the forces at play and use the laws of motion to find a solution.