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

[kuruman]

Thank you so much! That is a very nice guide. Using this I get that T_1 = m_1*(2*v^2/L-g)= -11.78 N (down), G_1 = m_1*g = -19.62 N (down), T_2 = m_2*(2*v^2/L+g) = 41.19 N, G_2 = m_2*g = -29.43 N. Do you agree with this?

I do not agree. Check your algebra.
I thought you were referring to the larger mass as $m_1$ as defined in my FBD in post #29.

 

[haruspex]

The acceleration is up towards the center.

$T_1$ corresponds to $m_1$, the lesser mass. Its acceleration will be downward.

 

[kuruman]

$T_1$ corresponds to $m_1$, the lesser mass. Its acceleration will be downward.

Yes, I was referring to my $m_1$ in he FBD in post #29 which is the larger mass and I thought that OP did too. I deleted the earlier post to avoid confusion. I think I will remove myself from cooking this broth any further.

 

[ymnoklan]

Thank you all so much for your help! I am still stuck with the hardest one though… the force on the axis? What do they ask for here? What is different from this and the previous question? And where do I even begin?

 

[Steve4Physics]

Thank you all so much for your help! I am still stuck with the hardest one though… the force on the axis? What do they ask for here? What is different from this and the previous question? And where do I even begin?

Can you solve some simpler problems first? These should lead you in the right direction.

An object, weight 10N, is attached to one end of a rod of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

Q1 When the mass is at the top, suppose |centripetal force| = 15N.
a) What is the force (size and direction) exerted by the rod on the mass?
b) What is the tension or compression in the rod?
c) What is the force (size and direction) exerted by the rod on the pivot?

Q2. Same again but this time the speed is slower so |centripetal force| = 7N.

Q3. Same again with mass at bottom and |centripetal force| = 15N.

Q4. Same again with mass at bottom and |centripetal force| = 7N.

When you’ve sorted these out, go back to the original problem, Pretend there are two rods – one rod (length L//2) connects $m_1$ to the pivot; the other rod (also length L/2) connects $m_2$ to the pivot. The force on the pivot is then the (vector) sum of the forces exerted by each rod on the pivot.

(I’m off to bed now so can’t contribute for several hours!)

 

[Lnewqban]

I am still stuck with the hardest one though… the force on the axis? What do they ask for here?

Copied from the problem statement: "You hold the bar in a horizontal position and release the system."

At the time that you release it, the system would free fall, unless a force at the axis is preventing that from happening.

As the center of mass of the system is located off that axis, a rotation, and related centrifugal forces, are induced.

While rotation happens, the centrifugal forces are directed radially, while the weight forces are directly downwards.

The question seems to be directed to the point at which the system has reached a vertical position (top mass-axis-bottom mass all aligned vertically).

 

[ymnoklan]

Can you solve some simpler problems first? These should lead you in the right direction.

An object, weight 10N, is attached to one end of a rod of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

Q1 When the mass is at the top, suppose |centripetal force| = 15N.
a) What is the force (size and direction) exerted by the rod on the mass?
b) What is the tension or compression in the rod?
c) What is the force (size and direction) exerted by the rod on the pivot?

Q2. Same again but this time the speed is slower so |centripetal force| = 7N.

Q3. Same again with mass at bottom and |centripetal force| = 15N.

Q4. Same again with mass at bottom and |centripetal force| = 7N.

When you’ve sorted these out, go back to the original problem, Pretend there are two rods – one rod (length L//2) connects $m_1$ to the pivot; the other rod (also length L/2) connects $m_2$ to the pivot. The force on the pivot is then the (vector) sum of the forces exerted by each rod on the pivot.

(I’m off to bed now so can’t contribute for several hours!)

Thank you for the examples! I am not sure if I understand them correctly though. Trying to solve them I get:
Q1: a) 5 N down, b) 5 N (equal in magnitude to a), c) 5 N up (opposite to a)
Q2: a) 3 N up, b) 3 N, c) 3 N down
Q3: a) 25 N up, b) 25 N, c) 25 N down
Q4: a) 17 N up, b) 17 N, c) 17 N down
Based on this I would guess that the force on the axis is then sqrt(T_1^2 + T_2^2) = sqrt(11.78^2 + 41.18^2) ≈ 42.84 N downwards. What do you think of this?

 

[ymnoklan]

Copied from the problem statement: "You hold the bar in a horizontal position and release the system."

At the time that you release it, the system would free fall, unless a force at the axis is preventing that from happening.

As the center of mass of the system is located off that axis, a rotation, and related centrifugal forces, are induced.

While rotation happens, the centrifugal forces are directed radially, while the weight forces are directly downwards.

The question seems to be directed to the point at which the system has reached a vertical position (top mass-axis-bottom mass all aligned vertically).

Hm okay, do you mean that the force on the axis will be the resultant force of the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
Then I would get this:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g)
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N

 

[Steve4Physics]

You're getting there!

Thank you for the examples! I am not sure if I understand them correctly though. Trying to solve them I get:
Q1: a) 5 N down, b) 5 N (equal in magnitude to a), c) 5 N up (opposite to a)

Agreed. Because the direction of the force of the rod on the mass (answer a)) is downwards and the the mass is at the top, this tells us that the rod is in tension, So the answer to part b) is really 'Tension = 5N'.

Q2: a) 3 N up, b) 3 N, c) 3 N down

Agreed. But the answer to part b) is really 'Compression = 3N'.

Q3: a) 25 N up, b) 25 N, c) 25 N down

Agreed. But the answer to part b) is really 'Tension= 25N'.

Q4: a) 17 N up, b) 17 N, c) 17 N down

Agreed. But the answer to part b) is really 'Tension = 17N'.

(Note that the rod must always be in tension when the mass is at the bottom position, but it can be in tension or compression when the mass is at the top position. Worth thinking about if not clear. )

Based on this I would guess that the force on the axis is then sqrt(T_1^2 + T_2^2) = sqrt(11.78^2 + 41.18^2) ≈ 42.84 N downwards. What do you think of this?

Disagree. Guessing is dangerous and no substitute for careful thinking!

When we add two forces by vector addition the answer depends on the angle between their directions. Look at these examples:

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 90$^0$ so we use Pythagoras and vector addition gives $\sqrt{2^2 +5^2}$ at an angle of $\tan^{-1} ( \frac 52)$ to the horizontal.

 

[ymnoklan]

You're getting there!


Agreed. Because the direction of the force of the rod on the mass (answer a)) is downwards and the the mass is at the top, this tells us that the rod is in tension, So the answer to part b) is really 'Tension = 5N'.


Agreed. But the answer to part b) is really 'Compression = 3N'.


Agreed. But the answer to part b) is really 'Tension= 25N'.


Agreed. But the answer to part b) is really 'Tension = 17N'.

(Note that the rod must always be in tension when the mass is at the bottom position, but it can be in tension or compression when the mass is at the top position. Worth thinking about if not clear. )


Disagree. Guessing is dangerous and no substitute for careful thinking!

When we add two forces by vector addition the answer depends on the angle between their directions. Look at these examples:

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 90$^0$ so we use Pythagoras and vector addition gives $\sqrt{2^2 +5^2}$ at an angle of $\tan^{-1} ( \frac 52)$ to the horizontal

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 900 so we use Pythagoras and vector addition gives 22+52 at an angle of tan−1⁡(52) to the horizontal.

I get that, but in my case I thought the only forces working are the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
This would result in:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g) at angle tan^-1(((m_1+m_2)*g)/((m_1+m_1)*v^2/r))
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N at angle theta ≈ 68.22 degrees
What do you think of this then?

 

[Steve4Physics]

I get that, but in my case I thought the only forces working are the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
This would result in:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g) at angle tan^-1(((m_1+m_2)*g)/((m_1+m_1)*v^2/r))
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N at angle theta ≈ 68.22 degrees
What do you think of this then?

It is unreadable (you need to learn to use LaTeX) and clearly wrong because it uses a tangent!

I think you have misunderstood. In Post #42 you were wrong to use Pythagoras because the 2 forces you were adding were not perpendicular.

At the end of Post #44 I was trying to explain your mistake by using some simple examples.

You need to add the two forces on the pivot. What are the magnitude and direction of each force on the pivot?

 

[ymnoklan]

It is unreadable (you need to learn to use LaTeX) and clearly wrong because it uses a tangent!

I think you have misunderstood. In Post #42 you were wrong to use Pythagoras because the 2 forces you were adding were not perpendicular.

At the end of Post #44 I was trying to explain your mistake by using some simple examples.

You need to add the two forces on the pivot. What are the magnitude and direction of each force on the pivot?

I find this very difficult to get my head around, so thank you a lot for your help. On the pivot I only really find T_1 = 11.78 N upwards and T_2 = 41.18 N upwards. Aren't these kind of "on top of each other" when the rod is vertical? I think this seems very odd though.

 

[Steve4Physics]

On the pivot I only really find T_1 = 11.78 N upwards and T_2 = 41.18 N upwards. Aren't these kind of "on top of each other" when the rod is vertical? I think this seems very odd though.

Your values are OK although I get 11.772N and 41.202N using a spreadsheet, so it seems you have introduced rounding errors at some point(s). But one (or both) of the direction(s) is (are) wrong.

Call the upper half of the rod A; this part connects the pivot to $m_1$.
Call the lower half of the rod B; this part connects the pivot to $m_2$.
You need to treat A and B as if they are separate rods connected to the same pivot.

What is the force (size and direction) exerted by A on the pivot?
What is the force (size and direction) exerted by B on the pivot?
(These questions are just like Q1-Q4 in Post #40 - which you correctly answered).

The total force on the pivot is then the sum of these 2 forces.

 

[ymnoklan]

Your values are OK although I get 11.772N and 41.202N using a spreadsheet, so it seems you have introduced rounding errors at some point(s). But one (or both) of the direction(s) is (are) wrong.

Call the upper half of the rod A; this part connects the pivot to $m_1$.
Call the lower half of the rod B; this part connects the pivot to $m_2$.
You need to treat A and B as if they are separate rods connected to the same pivot.

What is the force (size and direction) exerted by A on the pivot?
What is the force (size and direction) exerted by B on the pivot?
(These questions are just like Q1-Q4 in Post #40 - which you correctly answered).

The total force on the pivot is then the sum of these 2 forces.

I mean, obviously the sum of the forces must be towards the center of the "circle", which would mean that the total force by A would be downwards and the force by B would be upwards. Do you mean that the total force on the pivot and on the axis is 41.2 N - 11.8 N ≈ 29.4 N upwards (rounded) ?
If so, what are your thoughts on the directions on the forces in the rod then? I thought this would be T_1 ≈ 11.8 N up, G_1 ≈ 19.6 N down, T_2 ≈ 41.2 N up and G_2 ≈ 29.4 N down.
What do you think of this?

 

[Steve4Physics]

I mean, obviously the sum of the forces must be towards the center of the "circle"

No. The pivot IS the centre of the circle. The direction of the total force on the pivot can't be towards itself; that wouldn't make sense.

The centripetal force on a object is always towards the centre of the circular motion. But the force on the pivot is not the centripetal force. Please try these questions (similar to the Post #40 questions):
______

Q1. An object weighs 19.62N. It is attached to one end of a rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the top, |centripetal force on mass| = 7.848N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction (up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction (up/down)) exerted by the rod on the pivot?
______________

Q2. Another object weighs 29.43N. It is attached to one end of another rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the bottom, |centripetal force on mass| = 11.772N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction(up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction) exerted by the rod on the pivot?
___________

Q3 If the 2 systems (in Q1 and Q2) are on the same pivot, what is the total force on the pivot?

 

[ymnoklan]

No. The pivot IS the centre of the circle. The direction of the total force on the pivot can't be towards itself; that wouldn't make sense.

The centripetal force on a object is always towards the centre of the circular motion. But the force on the pivot is not the centripetal force. Please try these questions (similar to the Post #40 questions):
______

Q1. An object weighs 19.62N. It is attached to one end of a rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the top, |centripetal force on mass| = 7.848N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction (up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction (up/down)) exerted by the rod on the pivot?
______________

Q2. Another object weighs 29.43N. It is attached to one end of another rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the bottom, |centripetal force on mass| = 11.772N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction(up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction) exerted by the rod on the pivot?
___________

Q3 If the 2 systems (in Q1 and Q2) are on the same pivot, what is the total force on the pivot?

I don’t understand question d. What do you mean by compressive and tensile forces? Do you mean for example the gravitational force as a tensile force and the force by the rod as a compressive force?
Attempting at your questions I find:
Q1,
a) down, b) 11.772 N up, c) tension, d) -, e) 11.772 N down
Q2,
a) up, b) 41.202 N up, c) tension, d) -, e) 41.202 N down
Q3,
I then find 11.772 N + 41.202 N = 52.974 N down, but this seems odd? What do you think of my answers?

 

[Steve4Physics]

I don’t understand question d. What do you mean by compressive and tensile forces?

Sounds like you are missing some basic background information/knowledge. I'm not sure this is the correct place for Physics 101 but...

Consider a spring with open turns (gaps between adjacent coils).

If you stretch the spring (make it longer than its natural length) we say the spring is under tension - there is a tensile force inside it. Pull one end to the left with a force of 7N and the other end to the right with a force of 7N till equilibrium is reached. The tension (tensile force) in the spring is said to be 7N. (Note that the net force on the spring is zero.)

Similarly, if you compress the spring (make it shorter than its natural length) we say the spring is under compression. You could push each end in with a force of 7N to make the spring shorter; at equilibrium the compressive force is 7N.

This doesn't just apply to springs. If you have a rod, you could grab each end and try to stretch it in the same way as the spring. The rod is then under tension. It experiences a tensile force. It's length will increase but typically only by a very small amount.

Similarly if you push the rod's ends inwards, the rod is then under compression.

Try it for yourself. Grab one end of a pencil in your left hand and the other in your right hand. Pull outwards. The force exerted by each hand equals the tensile force - ask yourself: what force is the pencil exerting on each hand and what force does each hand exert on the pencil (Newton's 3rd law)? Repeat, but pushing the ends of the pencil inwards.

If you have objects X and Y at the end of a rod then:

- if the rod has a compressive force of 7N, it will exert a 7N outwards force on X and a 7N outwards force on Y, to try and push X and Y apart.

- if the rod has a tensile force of 7N it will exert a 7N inwards force on X and a 7N inwards force on Y, to try to pull X and Y together.

Once you've thought carefully about this, try the Post #50 questions again.

Edit - minor changes.

 

[ymnoklan]

Sounds like you are missing some basic background information/knowledge. I'm not sure this is the correct place for Physics 101 but...

Consider a spring with open turns (gaps between adjacent coils).

If you stretch the spring (make it longer than its natural length) we say the spring is under tension - there is a tensile force inside it. Pull one end to the left with a force of 7N and the other end to the right with a force of 7N till equilibrium is reached. The tension (tensile force) in the spring is said to be 7N. (Note that the net force on the spring is zero.)

Similarly, if you compress the spring (make it shorter than its natural length) we say the spring is under compression. You could push each end in with a force of 7N to make the spring shorter; at equilibrium the compressive force is 7N.

This doesn't just apply to springs. If you have a rod, you could grab each end and try to stretch it in the same way as the spring. The rod is then under tension. It experiences a tensile force. It's length will increase but typically only by a very small amount.

Similarly if you push the rod's ends inwards, the rod is then under compression.

Try it for yourself. Grab one end of a pencil in your left hand and the other in your right hand. Pull outwards. The force exerted by each hand equals the tensile force - ask yourself: what force is the pencil exerting on each hand and what force does each hand exert on the pencil (Newton's 3rd law)? Repeat, but pushing the ends of the pencil inwards.

If you have objects X and Y at the end of a rod then:

- if the rod has a compressive force of 7N, it will exert a 7N outwards force on X and a 7N outwards force on Y, to try and push X and Y apart.

- if the rod has a tensile force of 7N it will exert a 7N inwards force on X and a 7N inwards force on Y, to try to pull X and Y together.

Once you've thought carefully about this, try the Post #50 questions again.

Edit - minor changes.

Okay, so I think I get your point, and I think part of my problem is simply due to the language barrier (some physics terms can be quite difficult to translate). However I still struggle to see what’s wrong with my previous answers? Based on what you just wrote, I would think that the tensile force in Q1 is the force from the rod on the mass (pushing upwards), while in Q2 it is both the gravitational force and the force from the rod pulling down. Could you guide me on the right way? I feel like this is such a simple problem and I really want to understand it, but I must say I find it very difficult. Thank you for your help and patience!

 

[haruspex]

I would think that the tensile force in Q1 is the force from the rod on the mass (pushing upwards), while in Q2 it is both the gravitational force and the force from the rod pulling down.

In post #47, you found the forces the rod exerts on the masses.
By Newton’s 3rd law, the masses exert equal and opposite forces on the rod.
The axle also exerts a force on the rod.
These three forces are in balance, since the rod has no vertical acceleration.
Write that as an equation.

 

[Steve4Physics]

Okay, so I think I get your point, and I think part of my problem is simply due to the language barrier (some physics terms can be quite difficult to translate).

Yes, Fair enough.

However I still struggle to see what’s wrong with my previous answers? Based on what you just wrote, I would think that the tensile force in Q1 is the force from the rod on the mass (pushing upwards), while in Q2 it is both the gravitational fo N and rce and the force from the rod pulling down. Could you guide me on the right way? I feel like this is such a simple problem and I really want to understand it, but I must say I find it very difficult. Thank you for your help and patience!

Let’s call the section of rod between $m_1$ (at the top) and the pivot: ‘A’.

We know that the net force on $m_1$ is the centripetal force, 7.848N down. This force is the sum of:
a) $m_1$'s weight, 19.62N down and
b) the force of A on $m_1$, which is easily shown to be, 11.772N up.

Since the force of A on $m_1$ is 11.772N up, A must be in a compressed state; the compressive force is 11.772N.

(Alternatively, we could apply Newton's 3rd law and say that the force of $m_1$ on A is 11.772N down, compressing A.)

Since A is compressed (compressive force is 11.772N) it follows that the force of A on the pivot is 11.772N down. (A is like a compressed spring, pushing its ends outwards.)

The force exerted by the bottom half of the rod on the pivot can be found with similar logic. Then the two forces can be added.

The detailed working is best done using appropriate equations, but since it’s first necessary to have a conceptual understanding of what’s going on, I’ve keep the description fairly qualitative.

 

[ymnoklan]

Yes, Fair enough.


Let’s call the section of rod between $m_1$ (at the top) and the pivot: ‘A’.

We know that the net force on $m_1$ is the centripetal force, 7.848N down. This force is the sum of:
a) $m_1$'s weight, 19.62N down and
b) the force of A on $m_1$, which is easily shown to be, 11.772N up.

Since the force of A on $m_1$ is 11.772N up, A must be in a compressed state; the compressive force is 11.772N.

(Alternatively, we could apply Newton's 3rd law and say that the force of $m_1$ on A is 11.772N down, compressing A.)

Since A is compressed (compressive force is 11.772N) it follows that the force of A on the pivot is 11.772N down. (A is like a compressed spring, pushing its ends outwards.)

The force exerted by the bottom half of the rod on the pivot can be found with similar logic. Then the two forces can be added.

The detailed working is best done using appropriate equations, but since it’s first necessary to have a conceptual understanding of what’s going on, I’ve keep the description fairly qualitative.

Is the force at the bottom downwards then (because B is in tension)?
Resulting in 41.19 N - 11.772 N = 29.418 N down.

 

[Steve4Physics]

Is the force at the bottom downwards then (because B is in tension)?
Resulting in 41.19 N - 11.772 N = 29.418 N down.

No.

The upper half of the rod is 'A'. The lower half is 'B'.

Yes, B is in tension - how did you deduce that?

Say, you have a vertical spring in tension, held stretched between a support at the top end and a support at the bottom end:
- what direction is the force of the spring on the top-support?
- what direction is the force of the spring on the bottom-support?

 

[ymnoklan]

No.

The upper half of the rod is 'A'. The lower half is 'B'.

Yes, B is in tension - how did you deduce that?

Say, you have a vertical spring in tension, held stretched between a support at the top end and a support at the bottom end:
- what direction is the force of the spring on the top-support?
- what direction is the force of the spring on the bottom-support?

Yes, B is in tension - how did you deduce that?

Simply because B is at the bottom, therefore it must be in tension (while when it is at the top it can be either in tension or compressed based on the sizes of the gravitational force and the centripetal force I assume).

what direction is the force of the spring on the top-support?

Up?

what direction is the force of the spring on the bottom-support?

Down?

 

[Steve4Physics]

I asked:
"Say, you have a vertical spring in tension, held stretched between a support at the top end and a support at the bottom end:
- what direction is the force of the spring on the top-support?"
You said:

Up?

That's wrong!

I then asked:
"- what direction is the force of the spring on the bottom-support?"
You said:

Down?

That's wrong too!

Could this be a language problem? To rephrase th questions, I'm asking:
- what direction is the force exerted by the spring on the top-support? and
- what direction is the force exerted by the spring on the bottom-support?

All I can do is suggest you try this:

Get an elastic band.
Hold part of it with your left hand.
Hold the opposite side with your right hand.
Put your left hand above your right hand so the band is vertical.
Move your left hand up a bit and move your right hand down a bit so the band is stretched (in tension).

Ask yourself:
1. What force does my left hand exert on the band?
2. What force does the band exert on my left hand?
3. What force does my right hand exert on the band?
4. What force does the band exert on my right hand?

When you have tried this and thought about it, check your answers by clicking on the fuzzy spoiler below:

1. The left hand is pulling the band up. So the answer is up.
2. The band is pulling the left hand down. So the answer is down. If this is not clear, revise Newton's 3rd law!
3. The right hand is pulling the band down. So the answer is down.
4. The band is pulling the right hand up. So the answer is up. If this is not clear, revise Newton's 3rd law!

If you still don't get it, I'm at a loss how to explain it differently Others are welcome to chip-in!

Edit: typo'.

 

[ymnoklan]

I asked:
"Say, you have a vertical spring in tension, held stretched between a support at the top end and a support at the bottom end:
- what direction is the force of the spring on the top-support?"
You said:

That's wrong!

I then asked:
"- what direction is the force of the spring on the bottom-support?"
You said:

That's wrong too!

Could this be a language problem? To rephrase th questions, I'm asking:
- what direction is the force exerted by the spring on the top-support? and
- what direction is the force exerted by the spring on the bottom-support?

All I can do is suggest you try this:

Get an elastic band.
Hold part of it with your left hand.
Hold the opposite side with your right hand.
Put your left hand above your right hand so the band is vertical.
Move your left hand up a bit and move your right hand down a bit so the band is stretched (in tension).

Ask yourself:
1. What force does my left hand exert on the band?
2. What force does the band exert on my left hand?
3. What force does my right hand exert on the band?
4. What force does the band exert on my right hand?

When you have tried this and thought about it, check your answers by clicking on the fuzzy spoiler below:

1. The left hand is pulling the band up. So the answer is up.
2. The band is pulling the left hand down. So the answer is down. If this is not clear, revise Newton's 3rd law!
3. The right hand is pulling the band down. So the answer is down.
4. The band is pulling the right hand up. So the answer is up. If this is not clear, revise Newton's 3rd law!

If you still don't get it, I'm at a loss how to explain it differently Others are welcome to chip-in!

Edit: typo'.

- what direction is the force exerted by the spring on the top-support? and
- what direction is the force exerted by the spring on the bottom-support?

Ahh, I get it now (and I got your other questions at first try, so guess there's still some hope for me getting this!). But I must say I struggle seeing how this is relevant for my problem.

The total force on the pivot is then the sum of these 2 forces.

This I thought would be the sum of the centripetal forces. Is that what you mean?
Σ F = 11.772 N (up) - 7.848 N (down) = 3.924 N up
or is it simply the sum of the weights, but directed upwards to support the rod and the blocks:
Σ F = 19.62 N (down) + 29.42 N (down) = 49.05 N (down) => (Newton's 3rd law then gives 49.05 N upwards)

 

[Steve4Physics]

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.

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?

 

[ymnoklan]

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?

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).

 

[Steve4Physics]

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?

 

[ymnoklan]

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

 

[Steve4Physics]

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.

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?

 

[ymnoklan]

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

 

[Steve4Physics]

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.

 

[Steve4Physics]

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.