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?kuruman said:Being terrible at FBDs does not mean that you cannot do something to stop from "being terrible". Here is a tep-by-step recipe that I have used in my classes. It will see you through as long as you perform each step in order without omitting any step or doing things in your head. I have attached figures next to each step t show how the FBD develops. The method works best if you use symbols instead of numbers.
View attachment 3516191. Identify the system and draw it.
The system is the mass ##m_1=3~## kg at the bottom of the trajectory.
2. Identify all the pieces of the Universe that interact with the system and count them.
The pieces that interact with ##m_1## are the Earth that attracts it and the rod that is attached to it. This makes a total of two pieces.
View attachment 3516213. Draw one and only one arrow representing the force exerted by each of the items in (2). Label each arrow unambiguously.
See figure on right. The Earth exerts force ##m_1g## that is always down. The force exerted from the rod must be in a direction along the rod, i.e. vertical. It cannot be up because if it were, the mass will accelerate vertically down and not go around in a circle.
4. Choose a convenient coordinate system and draw it. Enclose your system, axes and forces in a View attachment 351623box with dotted lines.
Positive direction is up as indicated by the arrow labeled ##y##.
View attachment 3516245. Declare the acceleration. Draw an arrow outside the box indicating the direction of the acceleration. If you have chosen the coordinate axes wisely, the acceleration will be along one of the principal axes (x or y). Label the arrow "m"a, where "m" is the symbol you used for the mass in (1).
Mass ##m_1## is going around in a circle. Since all the force vectors in the diagram are along the vertical, the acceleration can only be along the vertical, up or down. Now the mass is going around in a circle centered directly above ##m_1##. The acceleration is centripetal (towards the center), hence its direction is up.
6. Add all the forces vectorially to obtain the net force, the left side in Newton's second law. Check that the vector sum of the arrows inside the dotted line gives a resultant vector in the direction of the "mass times acceleration" vector that you drew outside the dotted line in (5).
Anything up is positive and anything down is negative, so $$F_{\text{net}}=F_{\text{rod}}-m_1g.$$7. Set the net force vector equal to the mass times acceleration vector outside the FBD. Solve for what you are asked to find.
$$F_{\text{net}}=F_{\text{rod}}-m_1g=m_1a$$ Since the acceleration is centripetal, $$F_{\text{rod}}-m_1g=m_1v^2/(L/2).$$ You have the speed ##v## from a previous part, you need to solve the force exerted by the rod on the mass. The force exerted by the mass on the rod has the same magnitude and opposite direction to that.
If you are serious about improving your free body diagram skills, apply this step-by-step method to find the force exerted by the other mass on the rod and consider using it in the future.
Thank you so much! That is a very nice guide. Using your tips 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?haruspex said:##G_1## and ##G_2## are different. Maybe that was just a typo.
As @Steve4Physics mentioned, it is much better to work algebraically. Don't plug in numbers until you have to. So far you have ##v^2=\frac{(m_2-m_1)gL}{(m_2+m_1)}##.
You need to define which way is positive for your forces. It doesn’t matter what you choose as long as, for each one, you are consistent.
Say we take towards the axis as positive always. Then we have ##T_1+G_1=m_1\frac{v^2}{L/2}##, ##T_2-G_2=m_2\frac{v^2}{L/2}##.
What does that give you for the tensions?
Yes, ##T_1 = -11.78 N##, but remember that we are taking towards the axis as positive. So which way is a force of -11.78N acting?ymnoklan said:I get that T_1 = m_1*(2*v^2/L-g)= -11.78 N (down),
I get that, but that disagrees with my free-body diagram. Is T_1 really upwards? Do you agree with the other directions though?haruspex said:Yes, ##T_1 = -11.78 N##, but remember that we are taking towards the axis as positive. So which way is a force of -11.78N acting?
The free body diagram should show the forces acting in whichever direction you are choosing as positive. It doesn’t matter if the actual direction is the other way.ymnoklan said:I get that, but that disagrees with my free-body diagram.
Yes. It is rotating quite slowly.ymnoklan said:Is T_1 really upwards?
Yes.ymnoklan said:Do you agree with the other directions though?
ymnoklan said: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?
##T_1## corresponds to ##m_1##, the lesser mass. Its acceleration will be downward.kuruman said:The acceleration is up towards the center.
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.haruspex said:##T_1## corresponds to ##m_1##, the lesser mass. Its acceleration will be downward.
Can you solve some simpler problems first? These should lead you in the right direction.ymnoklan said: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?
Copied from the problem statement: "You hold the bar in a horizontal position and release the system."ymnoklan said:I am still stuck with the hardest one though… the force on the axis? What do they ask for here?
Thank you for the examples! I am not sure if I understand them correctly though. Trying to solve them I get:Steve4Physics said: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!)
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?Lnewqban said: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).
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'.ymnoklan said: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. But the answer to part b) is really 'Compression = 3N'.ymnoklan said:Q2: a) 3 N up, b) 3 N, c) 3 N down
Agreed. But the answer to part b) is really 'Tension= 25N'.ymnoklan said:Q3: a) 25 N up, b) 25 N, c) 25 N down
Agreed. But the answer to part b) is really 'Tension = 17N'.ymnoklan said:Q4: a) 17 N up, b) 17 N, c) 17 N down
Disagree. Guessing is dangerous and no substitute for careful thinking!ymnoklan said: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?
Steve4Physics said: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
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?Steve4Physics said: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.
It is unreadable (you need to learn to use LaTeX) and clearly wrong because it uses a tangent!ymnoklan said: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?
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 said: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?
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.ymnoklan said: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.
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) ?Steve4Physics said: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.
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.ymnoklan said:I mean, obviously the sum of the forces must be towards the center of the "circle"
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?Steve4Physics said: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?
Sounds like you are missing some basic background information/knowledge. I'm not sure this is the correct place for Physics 101 but...ymnoklan said:I don’t understand question d. What do you mean by compressive and tensile forces?
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!Steve4Physics said: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.
In post #47, you found the forces the rod exerts on the masses.ymnoklan said: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.
Yes, Fair enough.ymnoklan said: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).
Let’s call the section of rod between ##m_1## (at the top) and the pivot: ‘A’.ymnoklan said: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!
Is the force at the bottom downwards then (because B is in tension)?Steve4Physics said: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.
No.ymnoklan said: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 said: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?
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).Steve4Physics said:Yes, B is in tension - how did you deduce that?
Up?Steve4Physics said:what direction is the force of the spring on the top-support?
Down?Steve4Physics said:what direction is the force of the spring on the bottom-support?
That's wrong!ymnoklan said:Up?
That's wrong too!ymnoklan said:Down?
Steve4Physics said: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'.
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.Steve4Physics said:- 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?
This I thought would be the sum of the centripetal forces. Is that what you mean?Steve4Physics said:The total force on the pivot is then the sum of these 2 forces.