RODS, RINGS AND STRINGS isaac physics
This is my working for this question. https://isaacphysics.org/questions/rods_rings_string?board=21d09d04-feff-4743-9057-767c175c9131&stage=all
I believe I have to get F_spring? I am not sure if my calculations are on track or wether they are clean out of order. Any help would be appreciated.
I believe I have to get F_spring? I am not sure if my calculations are on track or wether they are clean out of order. Any help would be appreciated.
Reply 1
Original post by Javier García
This is my working for this question. https://isaacphysics.org/questions/rods_rings_string?board=21d09d04-feff-4743-9057-767c175c9131&stage=all
I believe I have to get F_spring? I am not sure if my calculations are on track or wether they are clean out of order. Any help would be appreciated.
I believe I have to get F_spring? I am not sure if my calculations are on track or wether they are clean out of order. Any help would be appreciated.
Reply 2
21
Original post by javier garcía
Not fully worked it through but cant help thinking youre approaching it in the "wrong" way, so setting up complex equations which are hard to solve rather than thinking about spotting simple relationships. A few things that come to mind
•
If the peg is smooth, is the tension in the string constant?
•
For the rod/ring if you resolve parallel/perp can you get T and trig(alpha) in terms of mg and trig(theta)
•
If you took moments for the rod/ring, youd bring Rs position into it
•
For the ring, parallel/perpendicular should give you N and ...
•
Rod will then bring the spring / position into it and get the spring constant
(edited 1 year ago)
Reply 3
Original post by mqb2766
Not fully worked it through but cant help thinking youre approaching it in the "wrong" way, so setting up complex equations which are hard to solve rather than thinking about spotting simple relationships. A few things that come to mind
•
If the peg is smooth, is the tension in the string constant?
•
If you take moments about ... can you get a simple expression for T (hint 4)
•
Similarly can you find the angle the other end of the string makes with the rod
•
Youll need get trig(theta) so can you come up with a simple relationship
Reply 4
21
To get sin(alpha) Id probably just have resolved normally so
T + Tsin(alpha) = 4/3 mg cos(theta)
Divide by T and used previous result
sin(alpha) = 5/3 - 1 = 2/3
Thinking about which is the simplest way to get such stuff is a key thing in questions like this.
Anyway, Youve used two equations for hint 4, so you try resolving along the bar and you should get a relationship for theta as now the weight will resovle as sin(theta) and the tension is ~cos(theta) so
Tcos(alpha) = 4/3 mg sin(theta)
tan(theta) = 3/5 cos(alpha)
and you can work out cos(alpha) using pythagoras/right triangle. You know then sin(theta), cos(theta) as well ...
As above for each equilibrium diagram think about resolving paralllel/perpendicular or horizontal/vertical and one moment and think about what it will tell you (or two moments and one direction).
T + Tsin(alpha) = 4/3 mg cos(theta)
Divide by T and used previous result
sin(alpha) = 5/3 - 1 = 2/3
Thinking about which is the simplest way to get such stuff is a key thing in questions like this.
Anyway, Youve used two equations for hint 4, so you try resolving along the bar and you should get a relationship for theta as now the weight will resovle as sin(theta) and the tension is ~cos(theta) so
Tcos(alpha) = 4/3 mg sin(theta)
tan(theta) = 3/5 cos(alpha)
and you can work out cos(alpha) using pythagoras/right triangle. You know then sin(theta), cos(theta) as well ...
As above for each equilibrium diagram think about resolving paralllel/perpendicular or horizontal/vertical and one moment and think about what it will tell you (or two moments and one direction).
(edited 1 year ago)
Reply 5
Original post by mqb2766
To get sin(alpha) Id probably just have resolved normally so
T + Tsin(alpha) = 4/3 mg cos(theta)
Divide by T and used previous result
sin(alpha) = 5/3 - 1 = 2/3
Thinking about which is the simplest way to get such stuff is a key thing in questions like this.
Anyway, Youve used two equations for hint 4, so you try resolving along the bar and you should get a relationship for theta as now the weight will resovle as sin(theta) and the tension is ~cos(theta) so
Tcos(alpha) = 4/3 mg sin(theta)
tan(theta) = 3/4 cos(alpha)
and you can work out cos(alpha) using pythagoras/right triangle. You know then sin(theta), cos(theta) as well ...
As above for each equilibrium diagram think about resolving paralllel/perpendicular or horizontal/vertical and one moment and think about what it will tell you (or two moments and one direction).
T + Tsin(alpha) = 4/3 mg cos(theta)
Divide by T and used previous result
sin(alpha) = 5/3 - 1 = 2/3
Thinking about which is the simplest way to get such stuff is a key thing in questions like this.
Anyway, Youve used two equations for hint 4, so you try resolving along the bar and you should get a relationship for theta as now the weight will resovle as sin(theta) and the tension is ~cos(theta) so
Tcos(alpha) = 4/3 mg sin(theta)
tan(theta) = 3/4 cos(alpha)
and you can work out cos(alpha) using pythagoras/right triangle. You know then sin(theta), cos(theta) as well ...
As above for each equilibrium diagram think about resolving paralllel/perpendicular or horizontal/vertical and one moment and think about what it will tell you (or two moments and one direction).
I have substituted in the value of alpha I got into the equations above, to get theta, then I tried to get T and since T=lambda, (as x and L are both l/2 they cancell out) I know therefore that lanbda=51.4N, but apparently is incorrect. (To be honest, I do not know where you get those equations from)
Reply 6
21
Original post by javier garcía
I have substituted in the value of alpha I got into the equations above, to get theta, then I tried to get T and since T=lambda, (as x and L are both l/2 they cancell out) I know therefore that lanbda=51.4N, but apparently is incorrect. (To be honest, I do not know where you get those equations from)
First one is resolve perpendicular to the whole system (hint 4) so
T + Tsin(alpha) = 4/3 mg cos(theta)
Second one is resolve parallel (along) the whole system (hint 4) so
Tcos(alpha) = 4/3 mg sin(theta)
That way you have 3 equations where you can get T, theta and alpha (or the trig values of them as it keeps the working as surds - not strictly necessary for the question).
Then plug into resolving parallel for the rod (hint 6) to get lambda.
Sometimes its worth taking a step back and thinking of the "bigger picture". So working back from hint 6, you need to know theta to get lambda. From hint 4, we see that the important info is T, theta and alpha and as we have 3 ways to determine equilibrium, we can determine them all in a relatively simple way. Each calculation / balance equation is relatively simple.
(edited 1 year ago)
Reply 7
Original post by mqb2766
First one is resolve perpendicular to the whole system (hint 4) so
T + Tsin(alpha) = 4/3 mg cos(theta)
Second one is resolve parallel (along) the whole system (hint 4) so
Tcos(alpha) = 4/3 mg sin(theta)
That way you have 3 equations where you can get T, theta and alpha (or the trig values of them as it keeps the working as surds - not strictly necessary for the question).
Then plug into resolving parallel for the rod (hint 6) to get lambda.
Sometimes its worth taking a step back and thinking of the "bigger picture". So working back from hint 6, you need to know theta to get lambda. From hint 4, we see that the important info is T, theta and alpha and as we have 3 ways to determine equilibrium, we can determine them all in a relatively simple way. Each calculation / balance equation is relatively simple.
T + Tsin(alpha) = 4/3 mg cos(theta)
Second one is resolve parallel (along) the whole system (hint 4) so
Tcos(alpha) = 4/3 mg sin(theta)
That way you have 3 equations where you can get T, theta and alpha (or the trig values of them as it keeps the working as surds - not strictly necessary for the question).
Then plug into resolving parallel for the rod (hint 6) to get lambda.
Sometimes its worth taking a step back and thinking of the "bigger picture". So working back from hint 6, you need to know theta to get lambda. From hint 4, we see that the important info is T, theta and alpha and as we have 3 ways to determine equilibrium, we can determine them all in a relatively simple way. Each calculation / balance equation is relatively simple.
Thanks, really appreciate it. So let us recap. Essentially in order to do the question we have to consider moments about any point you like, but points that would simplify things, for example, I took moments about the point where the ring stands, as that gave me an expression for T in terms of alpha, then you can take moments about the point in contact with the floor, to get another equation involving T, alpha and theta. Then you could resolve parallel to the rod to get another equation involving T,alpha and theta and you could resolve in the x direction to get another expression in terms of T, alpha and theta. This way, you have three equations or even 4, involving three unknowns. Then you could use the value of theta to resolve parallel to the rod in hint 6, as that involved F_spring and mg*sin(theta). I took a long time to do it, it is pretty hard.
(edited 1 year ago)
Reply 8
21
Original post by javier garcía
Thanks, really appreciate it. So let us recap. Essentially in order to do the question we have to consider moments about any point you like, but points that would simplify things, for example, I took moments about the point where the ring stands, as that gave me an expression for T in terms of alpha, then you can take moments about the point in contact with the floor, to get another equation involving T, alpha and theta. Then you could resolve parallel to the rod to get another equation involving T,alpha and theta and you could resolve in the x direction to get another expression in terms of T, alpha and theta. This way, you have three equations or even 4, involving three unknowns. Then you could use the value of theta to resolve parallel to the rod in hint 6, as that involved F_spring and mg*sin(theta). I took a long time to do it, it is pretty hard.
Sort of and you should have covered this in your maths class? Even though these questions are more complex, its still the same principle of thinking about what you need to find to determine a solution and picking the simplest (a simple) way to go. The stuff below has a few if/buts and ... but is some broad guidelines
•
For a single body (here the rod is different from the ring and both are different from the complete system, hence the hints in 4-6), you could form equilibrium equations in two directions and take moments about one point, so you can form 3 linearly independent equations and hence determine 3 things (or find 3 relationships if there are more than 3 unknowns).
•
The two usual directions are x-y or parallel-perpendicular, though you could take combinations of those. Pick ones that eliminate things you dont want to focus on.
•
You usually do 2 directions + 1 moment, though you could do 2 moments + 1 direction. Again pick the point you take moments about to eliminate things you dont want to focus on. You seemed to do 2 moments for the complete body, I suggested 1 moment + 2 directions as it seemed a bit simpler.
If its a more complex problem like this, sometimes its list the possible equilbrium equations (even roughly) and think about which are most useful. Some will be more complex to solve (simultaneously) and its easy to get lost with the numberof variables/unknowns in such problems. https://www.studysmarter.co.uk/explanations/math/mechanics-maths/rigid-bodies-in-equilibrium/
https://engineeringstatics.org/Chapter_05-2d-rigid-body-equilibrium.html
are ok, but Id google a few sources and make sure you understand the principles and why they choose certain ways. If you wrote up this one, its not that complex, though initially it the old problem of not being able to see the wood for the trees.
(edited 1 year ago)
Reply 9
Thanks for the web pages, they are useful material, I will try to read through them
Reply 10
Original post by mqb2766
Sort of and you should have covered this in your maths class? Even though these questions are more complex, its still the same principle of thinking about what you need to find to determine a solution and picking the simplest (a simple) way to go. The stuff below has a few if/buts and ... but is some broad guidelines
For this problem, the ring (hint 5) js largely ignored, though it was necessary to consider the direction of Tension and Normal.
If its a more complex problem like this, sometimes its list the possible equilbrium equations (even roughly) and think about which are most useful. Some will be more complex to solve (simultaneously) and its easy to get lost with the numberof variables/unknowns in such problems. https://www.studysmarter.co.uk/explanations/math/mechanics-maths/rigid-bodies-in-equilibrium/
https://engineeringstatics.org/Chapter_05-2d-rigid-body-equilibrium.html
are ok, but Id google a few sources and make sure you understand the principles and why they choose certain ways. If you wrote up this one, its not that complex, though initially it the old problem of not being able to see the wood for the trees.
•
For a single body (here the rod is different from the ring and both are different from the complete system, hence the hints in 4-6), you could form equilibrium equations in two directions and take moments about one point, so you can form 3 linearly independent equations and hence determine 3 things (or find 3 relationships if there are more than 3 unknowns).
•
The two usual directions are x-y or parallel-perpendicular, though you could take combinations of those. Pick ones that eliminate things you dont want to focus on.
•
You usually do 2 directions + 1 moment, though you could do 2 moments + 1 direction. Again pick the point you take moments about to eliminate things you dont want to focus on. You seemed to do 2 moments for the complete body, I suggested 1 moment + 2 directions as it seemed a bit simpler.
If its a more complex problem like this, sometimes its list the possible equilbrium equations (even roughly) and think about which are most useful. Some will be more complex to solve (simultaneously) and its easy to get lost with the numberof variables/unknowns in such problems. https://www.studysmarter.co.uk/explanations/math/mechanics-maths/rigid-bodies-in-equilibrium/
https://engineeringstatics.org/Chapter_05-2d-rigid-body-equilibrium.html
are ok, but Id google a few sources and make sure you understand the principles and why they choose certain ways. If you wrote up this one, its not that complex, though initially it the old problem of not being able to see the wood for the trees.
Point 4 of the latter website was useful for solving CYLINDER ROLLING DOWN A SLOPE, where I had to take moments about the point where the tension and the normal force intersected.
Reply 11
21
Original post by javier garcía
Point 4 of the latter website was useful for solving CYLINDER ROLLING DOWN A SLOPE, where I had to take moments about the point where the tension and the normal force intersected.
Agreed, and a similar thing can occur on the usual ladder/wall problems where the odd time it makes sense to take moments about where the wall meets the floor to eliminate both friction forces, instead of taking moments about the usual ends/middle of the ladder. Or even take moments where the two normal reactions to the wall and floor meet in order to eliminate them.
But more generally, its worth being systematic about listing which directions to resolve (x-y-parallel-perpendicular) so that youre perpendicular to unwanted forces and where to take moments so that unwanted forces have zero moments.
(edited 1 year ago)
Reply 12
Original post by mqb2766
Agreed, and a similar thing can occur on the usual ladder/wall problems where the odd time it makes sense to take moments about where the wall meets the floor to eliminate both friction forces, instead of taking moments about the usual ends/middle of the ladder. Or even take moments where the two normal reactions to the wall and floor meet in order to eliminate them.
But more generally, its worth being systematic about listing which directions to resolve (x-y-parallel-perpendicular) so that youre perpendicular to unwanted forces and where to take moments so that unwanted forces have zero moments.
But more generally, its worth being systematic about listing which directions to resolve (x-y-parallel-perpendicular) so that youre perpendicular to unwanted forces and where to take moments so that unwanted forces have zero moments.
Taking about unwanted forces, I tried doing just that in "Two Spheres" problem: https://isaacphysics.org/questions/two_spheres?stage=all
Since we do not want unwanted forces, I took moments about point C, as the video hints. However, since the point I am taking moments about is at a slight angle with respect to the horizontal, I have to "elongate the direction of the forces" and measure the perpendicular distance to the point as far as I am aware. I have taken moments in Sphere A and Sphere B. I would like to know wether the path I have ventured myself is somewhat good or if it is just a clean nonsense. This is my working. I realized I may have forgot to add 2mg*cos(theta) to the RHS of the equation when I simplified, but this is generally my approach.
(edited 1 year ago)
Reply 13
21
Original post by javier garcía
Taking about unwanted forces, I tried doing just that in "Two Spheres" problem: https://isaacphysics.org/questions/two_spheres?stage=all
Since we do not want unwanted forces, I took moments about point C, as the video hints. However, since the point I am taking moments about is at a slight angle with respect to the horizontal, I have to "elongate the direction of the forces" and measure the perpendicular distance to the point as far as I am aware. I have taken moments in Sphere A and Sphere B. I would like to know wether the path I have ventured myself is somewhat good or if it is just a clean nonsense. This is my working. I realized I may have forgot to add 2mg*cos(theta) to the RHS of the equation when I simplified, but this is generally my approach.
Since we do not want unwanted forces, I took moments about point C, as the video hints. However, since the point I am taking moments about is at a slight angle with respect to the horizontal, I have to "elongate the direction of the forces" and measure the perpendicular distance to the point as far as I am aware. I have taken moments in Sphere A and Sphere B. I would like to know wether the path I have ventured myself is somewhat good or if it is just a clean nonsense. This is my working. I realized I may have forgot to add 2mg*cos(theta) to the RHS of the equation when I simplified, but this is generally my approach.
In the middle of something but will look at it later, but a few things come to mind
•
Its fairly easy to get trig(theta) if you think about the centres of the spheres. Its the usual "trick" for circles in contact (used in am-gm).
•
Like T1=T2 in the previous question, can you use newton 3/force pairs to find any equal forces
•
Taking moments about C (sphere 1/sphere 2/both) will remove the contact forces
•
Taking moments about the centres (sphere 1/sphere 2) will relate the friction at C to the friction on the ground (roughly)
•
...
Note its better to start a new thread for a new question.
(edited 1 year ago)
Reply 14
You are correct, I should have started a new thread. I got part a (discarding the friction force, although I do not know why it is zero) and got the angle theta. For part b I need to find the minimum mu. I tried involving the friction force now and substitute in values for N and theta whilst solving moments about point C, but it is incorrect. Any help would be appreciated.
(edited 1 year ago)
Reply 15
21
Original post by Javier García
You are correct, I should have started a new thread. I got part a (discarding the friction force, although I do not know why it is zero) and got the angle theta. For part b I need to find the minimum mu. I tried involving the friction force now and substitute in values for N and theta whilst solving moments about point C, but it is incorrect. Any help would be appreciated.
For the friction being zero, you should be able to set up a couple of simple equations involving it and show it must be zero. Think about takimg moments ... and resolving ... though seeing your wokring for this and the other parts would help.
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