# How is the net force on the system equal to 0 initially?

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• kirito
kirito
TL;DR Summary
while solving a question about conservation of angular momentum, where the system initiates from a static state with net torque and net force is zero, I encountered difficulty identifying the the force opposing the block's motion , what maintains the block in place along the vertical axis opposing gravitational force

I understand that in the initial condition both the net torque and net force are zero since the system is in a static state , the net torque remains zero as the mass down is being pulled , the two blocks get pulled towards the axis of rotation by a radial force

but I am wondering what is keeping the block in place vertically opposing the gravitational force ? I understand that its in place because its attached so it has to do with the rod but the tension is perpendicular and acting in the x direction , what am I missing ?

Can you please give more details? If it's a schoolwork problem, post the exact problem statement, and upload any figures that are associated with the question. If it is a scenario that you made up, please make a more detailed drawing with everything labeled so we can help you figure out your Free Body Diagrams (FBDs) for this situation.

Lnewqban

In the system on the right, we have two masses, labeled as mass 1 and mass 2, connected by a rod and a rope. This setup allows the masses to slide closer when a third mass is pulled down. Initially, the system possesses an angular velocity.

for the system on the left, we have a rod with two masses fixed in place vertically. Due to these masses being fixed, there must be an upward force acting on them. However, tension and normal forces act along the x-axis. Does this arrangement imply that the rod is bent, even if not visibly apparent?"

kirito said:
for the system on the left, we have a rod with two masses fixed in place vertically.
What do you mean by "fixed vertically"?

Lnewqban said:
What do you mean by "fixed vertically"?
that there is no acceleration in the y direction

Lnewqban
kirito said:
In the system on the right, we have two masses, labeled as mass 1 and mass 2, connected by a rod and a rope. This setup allows the masses to slide closer when a third mass is pulled down. Initially, the system possesses an angular velocity.
Is it something like this what you are trying to describe?

kirito
Lnewqban said:
Is it something like this what you are trying to describe?
yes, particularly when the masses are on the edge just touching the bar sideways .

kirito said:
that there is no acceleration in the y direction
There is a reaction vertical force between the bar and the mass.
The weight of the mass is trying to bend the bar, but reaction forces inside the bar resist that.

kirito said:
yes, particularly when the masses are on the edge just touching the bar sideways .
There is a centrifugal effect on the masses when the central vertical bar is made to rotate.
The hand is forcing those masses to slide inwards under the pulling forces of the strings, which overcome the mentioned centrifugal forces.

kirito
Lnewqban said:
There is a reaction vertical force between the bar and the mass.
The weight of the mass is trying to bend the bar, but reaction forces inside the bar resist that.
I think what confuses me is the mass being hallow on the inside. When the bar is in contact with the upper part of the mass, it applies a force upwards. However, when it touches the lower part, it applies a force downwards. Wouldn't this normal force on the mass be totalled as zero ? meaning something else is opposing the gravitational force

kirito said:
I think what confuses me is the mass being hallow on the inside. When the bar is in contact with the upper part of the mass, it applies a force upwards. However, when it touches the lower part, it applies a force downwards. Wouldn't this normal force on the mass be totalled as zero ? meaning something else is opposing the gravitational force
Imagine that you are holding a pipe with your hand and strongly grabbing its whole perimeter: in this case, there will be many radial vectors of the many normal forces from your hand. Each of those force vectors get cancelled by its opposite one, resulting in a zero net force.

Now imagine the same, but simultaneously you are hanging your whole weight from the same pipe. Would your hand feel that "when it touches the lower part, it applies a force downwards"?

kirito
Lnewqban said:
Imagine that you are holding a pipe with your hand and strongly grabbing its whole perimeter: in this case, there will be many radial vectors of the many normal forces from your hand. Each of those force vectors get cancelled by its opposite one, resulting in a zero net force.

Would your hand feel that "when it touches the lower part, it applies a force downwards"?
I understood the first part (great visualisation helped a lot ), I just can't seem to imagine hanging my whole weight from the pipe (as in how )

kirito said:
... I just can't seem to imagine hanging my whole weight from the pipe (as in how )

kirito
Lnewqban said:
this was really informative thank you for the patience

kirito said:
this was really informative thank you for the patience
You are very welcome.
I have drawn the real-world situation between your sliding mass and the bar:

kirito

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