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

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The discussion revolves around the forces acting on a rotating rod with attached masses, specifically addressing the downward and upward forces experienced by the rod. Participants clarify the distinction between the forces acting on the rod and those exerted by the rod on the masses. The calculations for gravitational potential energy changes and centripetal forces are debated, with emphasis on the need for accurate free-body diagrams to identify all acting forces. There is confusion regarding the "force on the axis," which is linked to the net forces from the masses and their interactions with the rod. Overall, the thread emphasizes the importance of detailed calculations and clear representations of forces in solving physics problems.
  • #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.
 
  • #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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