Are there two downward forces acting on the pivot of the rotating rod?

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Homework Help Overview

The discussion revolves around a physics problem involving a rotating rod, forces acting on it, and the concepts of gravitational potential energy and centripetal force. Participants are exploring the relationships between the forces acting on the rod and the axis, as well as the implications of energy conservation in the system.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the meaning of "the force on the axis" and how it differs from the forces acting on the rod. There are discussions about the gravitational forces of the masses and their contributions to the overall force system. Some participants are questioning the calculations related to potential energy changes and the implications of mass displacements.

Discussion Status

The discussion is active, with participants providing insights and asking for clarifications on various aspects of the problem. Some guidance has been offered regarding the need for free-body diagrams and the importance of detailing calculations. Multiple interpretations of the forces and energy changes are being explored, indicating a productive exchange of ideas.

Contextual Notes

There are constraints regarding the clarity of notes shared by participants, which may hinder understanding. Additionally, the problem involves specific assumptions about the masses and their movements, which are under scrutiny.

  • #61
ymnoklan said:
This I thought would be the sum of the centripetal forces. Is that what you mean?
If by 'This' you mean the force on the pivot, then no. The centripetal forces do not act on the pivot, they are forces acting on the rotating masses.

ymnoklan said:
or is it simply the sum of the weights, but directed upwards to support the rod and the blocks:
No. The force on the pivot is not the sum of the weights (unless the system is at rest).

Answer these questions please.

From earlier posts we know the that:
- the upper half of the rod (A) is being compressed (compressive force = 11.772N);
- the lower half of the rod (B) is being stretched (tension =41.202N).

Q1 What is the size and direction of the force that A exerts on the pivot (if you are unsure, make sure you understand Post #59).

Q2 What is the size and direction of the force that B exerts on the pivot (if you are unsure, make sure you understand Post #59).

Q3. What is the total force on the pivot?
 
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  • #62
Steve4Physics said:
If by 'This' you mean the force on the pivot, then no. The centripetal forces do not act on the pivot, they are forces acting on the rotating masses.


No. The force on the pivot is not the sum of the weights (unless the system is at rest).

Answer these questions please.

From earlier posts we know the that:
- the upper half of the rod (A) is being compressed (compressive force = 11.772N);
- the lower half of the rod (B) is being stretched (tension =41.202N).

Q1 What is the size and direction of the force that A exerts on the pivot (if you are unsure, make sure you understand Post #59).

Q2 What is the size and direction of the force that B exerts on the pivot (if you are unsure, make sure you understand Post #59).

Q3. What is the total force on the pivot?
IMG_1985.jpeg

Q1: 11.772 N up, Q2: 41.202 N up, Q3: 11.772 N up + 41.202 N up = 85.974 N up
Is this correct?
(I added a sketch of my understanding of the situation).
 
  • #63
ymnoklan said:
View attachment 351715
Q1: 11.772 N up, Q2: 41.202 N up, Q3: 11.772 N up + 41.202 N up = 85.974 N up
Is this correct?
(I added a sketch of my understanding of the situation).
No. But the sketch was a good idea.

A is in a compressed state. That means it is pushing whatever is attached to its ends outwards. Like a compressed spring pushes whatever is attached to its ends outwards.

B is in tension. That means it is pulling whatever is attached to its ends inwards. Like a stretched spring pulls whatever is attached to its ends inwards.

I think you are confusing the forces which are applied to the rod with the forces applied by the rod.

Suppose objects X and Y are attached to a horizontal spring:
X\ \ \ \ \ \ \ \ \Y

If X and Y are pushed inwards, the spring is compressed:
##\rightarrow##\\\\\\\\\##\leftarrow## where the arrows show the forces of X and Y on the spring.

Then the force on X is outwards, to the left:
X##\leftarrow## where the arrow shows the force exerted by the spring on X.

And the force on Y is outwards, to the right:
##\rightarrow##Y where the arrow shows the force exerted by the spring on Y.

I used Newton's 3rd law.

A similar argument can be used if the spring is stretched.

Does that change your answers to the Post #61 questions?
 
  • #64
Steve4Physics said:
No. But the sketch was a good idea.

A is in a compressed state. That means it is pushing whatever is attached to its ends outwards. Like a compressed spring pushes whatever is attached to its ends outwards.

B is in tension. That means it is pulling whatever is attached to its ends inwards. Like a stretched spring pulls whatever is attached to its ends inwards.

I think you are confusing the forces which are applied to the rod with the forces applied by the rod.

Suppose objects X and Y are attached to a horizontal spring:
X\ \ \ \ \ \ \ \ \Y

If X and Y are pushed inwards, the spring is compressed:
##\rightarrow##\\\\\\\\\##\leftarrow## where the arrows show the forces of X and Y on the spring.

Then the force on X is outwards, to the left:
X##\leftarrow## where the arrow shows the force exerted by the spring on X.

And the force on Y is outwards, to the right:
##\rightarrow##Y where the arrow shows the force exerted by the spring on Y.

I used Newton's 3rd law.

A similar argument can be used if the spring is stretched.

Does that change your answers to the Post #61 questions?
That makes sense. Is this correct then?:
A is exerting a force outwards (down), B is exerting a force inwards (up) =>
ΣF = 41.202 N (by B) - 11.772 N (by A) = 29.43 N up
 
  • #65
ymnoklan said:
A is exerting a force outwards (down)
A exerts two forces: a force on the pivot and a force on ##m_1##. You need to be clear which force is down. But I assume you mean the force on the pivot, so you are correct.

ymnoklan said:
B is exerting a force inwards (up) =>
B is exerting inwards forces on the objects attached to its ends. But the pivot is at the top of B, so what is the direction of the force of B on the pivot?
 
  • #66
Steve4Physics said:
A exerts two forces: a force on the pivot and a force on ##m_1##. You need to be clear which force is down. But I assume you mean the force on the pivot, so you are correct.


B is exerting inwards forces on the objects attached to its ends. But the pivot is at the top of B, so what is the direction of the force of B on the pivot?
Of course. Is the pivot subjected to two downward forces then?
ΣF = 41.202 N (down by B) + 11.772 N (down by A) = 52.974 N down
 
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  • #67
Just in case there is any confusion...

We have objects X and Y attached to the ends of a rod.

When we say 'inwards' forces act on X and Y, we mean the forces on X and Y are directed towards the centre of the rod.

When we say 'outwards' force act on X and Y, we mean the forces on X and Y are directed away from the centre of the rod.

'Inwards' and 'outwards' in this context do not refer to towards and away from the centre of the circle.
 
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  • #68
ymnoklan said:
Of course. Is the pivot subjected to two downward forces then?
ΣF = 41.202 N (down by B) + 11.772 N (down by A) = 52.974 N down
Yes. You got it!
Edit: But don't forget to round the final answer to a suitable number of significant figures.
 

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