Hard kinematics problem: Block pushed onto a moving conveyor belt

In summary: I think you have a sign error there. A component of the frictional force should have the opposite sign to the corresponding component of the...
  • #36
IamVector said:
I think I need to learn more before solving these problems thank you for your efforts btw it will be a great help if you could kindly tell me your approach of doing maybe it will work out for me or you could explain me the hint provided and logic behind it I am really confused with that .
You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.
 
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  • #37
jbriggs444 said:
You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.
I think my concepts are a bit weak this ques is from jaan kalda kinematics pdf I tried doing it to make my concepts strong to reach level of Olympiads question but I managed to do only first 5 rest are out of my range thanks for your help.
 
  • #38
jbriggs444 said:
You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.
btw any advice from your side on improving my conceptual knowledge will be a great help
 
  • #39
IamVector said:
I think my concepts are a bit weak this ques is from jaan kalda kinematics pdf I tried doing it to make my concepts strong to reach level of Olympiads question but I managed to do only first 5 rest are out of my range thanks for your help.

This is not an easy question. There are two things. One is notational technique. Look at the way that @jbriggs444 understood my notation and why I had used the variables I did. That's something to work on.

The other thing is not needing a one-step solution. Easy mechanics questions are a bit like doing chess puzzles where you need to find checkmate in 1-2 moves. A problem like this is more like finding a checkmate in 4-5 moves. You need to keep going with a line of thought, analysis and mathematics and see where it leads. And, judge when it's not leading anywhere and backtrack.

Both of these things require practice and a certain discipline to stick with an idea and think it through.
 
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  • #40
PeroK said:
This is not an easy question. There are two things. One is notational technique. Look at the way that @jbriggs444 understood my notation and why I had used the variables I did. That's something to work on.

The other thing is not needing a one-step solution. Easy mechanics questions are a bit like doing chess puzzles where you need to find checkmate in 1-2 moves. A problem like this is more like finding a checkmate in 4-5 moves. You need to keep going with a line of thought, analysis and mathematics and see where it leads. And, judge when it's not leading anywhere and backtrack.

Both of these things require practice and a certain discipline to stick with an idea and think it through.
thanks I will try it one more time
 
  • #41
IamVector said:
thanks I will try it one more time
A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.

That said, I suspect there may be a very neat, quick solution to this. But, I don't see it.
 
  • #42
PeroK said:
A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.
I may be streamlining out some steps with my intuition leading me helter skelter to a nice diagram. But the geometric approach allows a result to be read off almost immediately.
 
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  • #43
PeroK said:
A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.

That said, I suspect there may be a very neat, quick solution to this. But, I don't see it.
ok thinking on it
 
  • #44
jbriggs444 said:
I may be streamlining out some steps with my intuition leading me helter skelter to a nice diagram. But the geometric approach allows a result to be read off almost immediately.
what is your approach??
 
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  • #45
IamVector said:
what is your approach??
Look at post #20 in this thread.

Following the approach means being able to understand that shifting from the belt frame to the ground frame amounts to adding a fixed rightward velocity (1 m/s in this case) to everything.
 
  • #46
jbriggs444 said:
Look at post #20 in this thread.

Following the approach means being able to understand that shifting from the belt frame to the ground frame amounts to adding a fixed rightward velocity (1 m/s in this case) to everything.
as you said to convert into lab frame I have to add 1m/s to everything I added it to my eq which will be equal to my velocity from lab frame(i think)
 
  • #47
IamVector said:
as you said to convert into lab frame I have to add 1m/s to everything I added it to my eq which will be equal to my velocity from lab frame(i think)
Can we see what you have done so that we can critique and possibly guide?
 
  • #48
It's become a contest bewteen inspiration and geometric insight on the one hand and honest algebraic toil on the other!
 
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  • #49
PeroK said:
It's become a contest bewteen inspiration and geometric insight on the one hand and honest algebraic toil on the other!
I think I got the answer! (μgt/√5 )i+(2-2μgt/√5)j got this equation after adding 1 then took its mod thus forming a quadratic in μgt now to find the min value of vector we need min value of quadratic so I find its vertex and took out its min value ie (4/5)^1/2 = (2)/(5^1/2)
 
  • #50
IamVector said:
I think I got the answer! (μgt/√5 )i+(2-2μgt/√5)j got this equation after adding 1 then took its mod thus forming a quadratic in μgt now to find the min value of vector we need min value of quadratic so I find its vertex and took out its min value ie (4/5)^1/2 = (2)/(5^1/2)
##\frac 2 {\sqrt 5} m/s## is the answer I get.

In general the minimum speed is ##v_b \cos \theta'##, where ##v_b## is the speed of the belt and ##\theta'## is the angle of motion relative to the belt.
 
  • #51
PeroK said:
##\frac 2 {\sqrt 5} m/s## is the answer I get.

In general the minimum speed is ##v_b \cos \theta'##, where ##v_b## is the speed of the belt and ##\theta'## is the angle of motion relative to the belt.
can you please explain further will be a great help
 
  • #52
PeroK said:
##\frac 2 {\sqrt 5} m/s## is the answer I get.

In general the minimum speed is ##v_b \cos \theta'##, where ##v_b## is the speed of the belt and ##\theta'## is the angle of motion relative to the belt.
is my approach right just want to confirm??
 
  • #53
IamVector said:
can you please explain further will be a great help
That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.
 
  • #54
PeroK said:
That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.
is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong
 
  • #55
IamVector said:
is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong
I haven't seen your approach.
 
  • #56
PeroK said:
I haven't seen your approach.
ok , so i got this equation (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j so to convert it into lab frame or ground frame I added 1m/s ie velocity of belt then I took the mod of the resultant vector ie (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j which in return formed a quadratic in μgt now we were asked to find the minimum vector so it will be minimum when the value of quadratic will be minimum so by differentiating i found the min val to be 4/5 thus on sqrt gives 2/(5^1/2)
 
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  • #57
PeroK said:
I haven't seen your approach.
I am preparing for Olympiads a word of advice will be appreciated and are the level of difficulty of questions I am doing sufficient??
 
  • #60
IamVector said:
Is it g/2? I am not sure
That's a quite a hard problem I think. You can't just guess that the answer is ##g/2## or ##3g/13## or whatever. You have to work it out.
 
  • #61
PeroK said:
That's a quite a hard problem I think. You can't just guess that the answer is ##g/2## or ##3g/13## or whatever. You have to work it out.
Well I took the component of mg acting on rod but I doubt it's a right approach.
 

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