How to Solve Dynamics Exercise Involving Force Representation?

  • #1
srnixo
47
10
Homework Statement
Actually, I solved half of the exercise, but i couldn't solve question 4 and 5. help me please, even with numerical application.

For example: in question 4 when calculating the ∆x = T/K , i dunno if T we take it as 4N or ( T=F⁰ =5N ) or even (T=F'=8N)

And about question 5, i dunno even how to start.
Relevant Equations
Newton's second law.
This is the exercise:

Please help me ( question 4 and 5).

IMG_20231213_152945.jpg

Here is my effort:
First, I represented the forces on both objects.

IMG_20231216_185345.jpg

Then, i found F⁰ = 5N (question 1)
IMG_20231216_185440.jpg

After that, (question 2) + (question 3)
IMG_20231216_185541.jpg

IMG_20231216_185643.jpg


IMG_20231216_185739.jpg


I hope it's even correct.
 
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  • #2
Would it be possible for you to concisely give the answers you get for each question you are in doubt about?
 
  • #3
erobz said:
Would it be possible for you to concisely give the answers you get for each question you are in doubt about?
I said: I tried all the questions, but I was not able to solve the fourth and fifth questions even though I had the numerical values from the above.

If anyone could help me , it would be great.
 
  • #4
srnixo said:
I said: I tried all the questions, but I was not able to solve the fourth and fifth questions even though I had the numerical values from the above.

If anyone could help me , it would be great.
for question 4. lets take the coordinate ##x_1## to be the displacement of block 1. Just before block 2 starts to move what are the forces acting on block 2 in the ##x## direction. Draw a simple diagram.
 
  • #5
erobz said:
for question 4. lets take the coordinate ##x_1## to be the displacement of block 1. Just before block 2 starts to move what are the forces acting on block 2 in the ##x## direction. Draw a simple diagram.
For block B: in the xx' we have:
Xx': T - Cbx' = 0 which means T=Cbx
In question 4: Block B is not in motion at all, so we were asked in the question to calculate ∆x.

+ ( I know that we calculate ∆x by the formula ∆x= T/K where K=200N/m, but i dunno the exact value of T.)
+ { Here my ability to complete the rest of the exercise ended )
 
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  • #6
Before block 2 moves the spring compresses by amount ##x_1##
 
  • #7
erobz said:
Before block 2 moves the spring compresses by amount ##x_1##
Yep which is Force T = Tension, i know this.
 
  • #8
srnixo said:
Yep which is Force T = Tension, i know this.
Ok, then write Newtons 2nd for block 1.
 
  • #9
erobz said:
Ok, then write Newtons 2nd for block 1.
And then!!!
 
  • #10
Scratch that, we know the displacement ##x_1## ( max displacement of block 1 before block 2 moves). We know the value of ##T## as you call it. Write the force balance on 2 in terms of the known parameters...
 
  • #11
erobz said:
Scratch that, we know the displacement ##x_1##. We know the value of ##T## as you call it. Write the force balance on 2 in terms of the known parameters...
I can directly calculate it by using ∆x=T/K , no need for equations, it's already the same , but i dunno the value of T , there are 3 possibilities which are T=4N or T=F⁰=5N or T=F'=8N, you need to well understand the situation.
 
  • #12
srnixo said:
I can directly calculate it by using ∆x=T/K , no need for equations, it's already the same , but i dunno the value of T , there are 3 possibilities which are T=4N or T=F⁰=5N or T=F'=8N, you need to well understand the situation.
The only forces acting on block 2 are the spring force and static friction...
 
  • #13
erobz said:
The only forces acting on block 2 are the spring force and static friction...
I'm truly confused , could you please provide me the solution directly if u're able :(
 
  • #14
srnixo said:
I'm truly confused , could you please provide me the solution directly if u're able :(
We aren't allowed to provide a solution until you have successfully solved the problem, by taking the advice given to you.

There are only 2 forces acting on block 2. The spring force pushing block 2 to the right ( the spring is in equilibrium as its being compressed - it has no mass - do a force balance on the spring if you have doubts ), and static friction is resisting motion of block 2...pushing to the left.
 
  • #15
erobz said:
There are only 2 forces acting on block 2. The spring force pushing block 2 to the right ( the spring is in equilibrium as its being compressed - it has no mass ), and static friction is resisting motion of block 2...pushing to the left.
I know this information,The problem is that I couldn't solve it. I understood all the exercise but i couldn't solve. I have writing problems and calculation issues
 
  • #16
srnixo said:
I know this information,The problem is that I couldn't solve it. I understood all the exercise but i couldn't solve. I have writing problems and calculation issues
Then what is ##x_1##? The displacement of block 1, just before block 2 begins to move? Use the force balance on block 2.
 
  • #17
erobz said:
Then what is ##x_1##? The displacement of block 1, just before block 2 begins to move? Use the force balance on block 2.
It is written that in ∆x=2cm block A starts moving with constant speed but B doesn't
Then in question 4: was all about ∆x of block B to start moving
 
  • #18
srnixo said:
It is written that in ∆x=2cm block A starts moving with constant speed but B doesn't
Then in question 4: was all about ∆x of block B to start moving
I think it's written that over range ##\Delta x = 2 \rm{cm}##, the block A moves with constant velocity. Thats just letting you know what the force balance on block 1 over that range is.

Put that aside for a moment and figure out how far the spring would need to be compressed before block 2 begins to move. Write the force balance on block two in terms of ##x_1##, and other parameters ##\mu_s, k, M, g ##
 
  • #19
erobz said:
I think it's written that over range ##\Delta x = 2 \rm{cm}##, the block A moves with constant velocity. Thats just letting you know what the force balance on block 1 over that range is.

Put that aside for a moment and figure out how far the spring would need to be compressed before block 2 begins to move. Write the force balance on block two in terms of ##x_1##, and other parameters ##\mu_s, k, M, g ##
Isn't 4N huh?
 
  • #20
srnixo said:
Isn't 4N huh?
No, ##T## is not ##4 \rm{N}##.

##x_1##?
 
  • #21
erobz said:
No, ##T## is not ##4 \rm{N}##.

##x_1##?
Then it's 5N , i considered the F⁰ case.
 
  • #22
srnixo said:
Then it's 5N , i considered the F⁰ case.
Yes, its static friction acting on block 2...it's not moving.
 
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  • #23
Yep yep 🤭 what about question 5 mr. Do u know anything about it! I actually didn't understand anything about it
 
  • #24
srnixo said:
Yep yep 🤭 what about question 5 mr. Do u know anything about it! I actually didn't understand anything about it
What did you get for ##x_1##? Something should stick out at you when comparing it to the range where block 1 moves at constant velocity ##\Delta x ##.
 
  • #25
erobz said:
What did you get for ##x_1##?
Bruh i wanna cry for real, if u're talking about question 4, i just understood the situation and used ∆x=5N\200 N/m. So simple 😭😭
 
  • #26
srnixo said:
Bruh i wanna cry for real, if u're talking about question 4, i just understood the situation and used ∆x=5N\200 N/m. So simple 😭😭
Bruh...you want to cry... This is like pulling my own teeth. I'm out, Good luck.
 
  • #27
erobz said:
Bruh...you want to cry... This is like pulling my own teeth. I'm out, Good luck.
Haha, at least thank you for giving me a hint about the tension force. I hope i find who will help me with question 5.
 
  • #28
srnixo said:
Haha, at least thank you for giving me a hint about the tension force. I hope i find who will help me with question 5.
I can pitch in. Start by drawing two free diagrams, one for each block, just before block B begins to move. Draw it to scale and point out which forces have equal magnitudes. Please post your diagram right side up.
 
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  • #29
First, a nitpick: you keep writing "tension", but the spring is under compression.

srnixo said:
about question 4, i just understood the situation and used ∆x=5N\200 N/m. So simple 😭😭
So the situation is now the same as in q3 but with a different displacement.
 
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  • #30
srnixo said:
Haha, at least thank you for giving me a hint about the tension force. I hope i find who will help me with question 5.
Pro tip: If you come here for help, try not to arrogantly resist every question/task that is asked of you. We aren't paid for our service. People are trying to help you learn how to solve the problem on their own time for their own reasons. Trust me, it's the more difficult route for both parties, but it's the most beneficial for you.
 
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  • #31
erobz said:
Pro tip: If you come here for help, try not to arrogantly resist every question/task that is asked of you. We aren't paid for our service. People are trying to help you learn how to solve the problem on their own time for their own reasons. Trust me, it's the more difficult route for both parties, but it's the most beneficial for you.
I am nothing but a student of science, and a student of the most difficult sciences, physics, and I am studying for my first university degree, so it is natural that I need help, and I am here to ask if anyone can help me and simplify things for me, not to make them more difficult. In addition, it is not a requirement that I help someone in exchange for money. If I have information, I will be happy to save someone from drowning in problems related to science. On top of all this, we are human beings with different abilities. Perhaps you are a person who is quick to calculate and understand quickly, but I am not necessarily like you. Perhaps I am a person of weak understanding.

[+ I am one of those people who, if they see the solution directly, take a certain amount of time to find the steps of this solution. It is not a requirement that finding the solution directly is a bad thing and makes the student lazy.]

If you want to help, do it with conscience and good intentions. Do not incite people to deal with others in the same way you do. Every human being has a certain intention and certain principles.Thank you again.
 
  • #32
haruspex said:
First, a nitpick: you keep writing "tension", but the spring is under compression.So the situation is now the same as in q3 but with a different displacement.
Alright alright thank you :)
 
  • #33
kuruman said:
I can pitch in. Start by drawing two free diagrams, one for each block, just before block B begins to move. Draw it to scale and point out which forces have equal magnitudes. Please post your diagram right side up.
Alright, i'm sorry my english is not that good for that reason i didn't understand the question.

If B hasn't moved yet, does this mean he has only two forces? ( The weight and contact force?)
 
  • #34
srnixo said:
I am nothing but a student of science, and a student of the most difficult sciences, physics, and I am studying for my first university degree, so it is natural that I need help, and I am here to ask if anyone can help me and simplify things for me, not to make them more difficult. In addition, it is not a requirement that I help someone in exchange for money. If I have information, I will be happy to save someone from drowning in problems related to science. On top of all this, we are human beings with different abilities. Perhaps you are a person who is quick to calculate and understand quickly, but I am not necessarily like you. Perhaps I am a person of weak understanding.

[+ I am one of those people who, if they see the solution directly, take a certain amount of time to find the steps of this solution. It is not a requirement that finding the solution directly is a bad thing and makes the student lazy.]

If you want to help, do it with conscience and good intentions. Do not incite people to deal with others in the same way you do. Every human being has a certain intention and certain principles.Thank you again.
Before progress can be made, both the helper and helpee must get on the same page. This requires cooperation with the helper. For instance, if they ask for diagrams give them the diagram. If they ask you to calculate something - calculate it - no matter how trivial it may seem. You may be taking a few steps backward, before you can move forward so the helper can access what you actually know and where a potential issue may be. There are many talented educators here - not saying I'm one of them - but you must trust the process they created and cooperate if you expect to learn.
 
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  • #35
erobz said:
Before progress can be made, both the helper and helpee must get on the same page. This requires cooperation with the helper. For instance, if they ask for diagrams give them the diagram. If they ask you to calculate something - calculate it - no matter how trivial it may seem. You may be taking a few steps backward, before you can move forward so the helper can access what you actually know and where a potential issue may be. There are many talented educators here - not saying I'm one of them - but you must trust the process they created and cooperate if you expect to learn.
He asked me about the diagram, i dunno how to draw it cuz i didn't even understand the situation of the question, for that reason i asked him about the forces of block B. But, What do you think I should do in this case? give up? Or should I tear up the exercise paper because I couldn't understand it?

I don't really know what your problem is with me, but not all students are treated the same way, each person has his own way of understanding. + I'm trying my best to do everything by myself. And let you know, I also don't have the time. I still have to study mathematics, chemistry, and Computer science as well. So, if somebody can help me to understand just the last question without making it look harder than it is ,would be nice and humane too.

Thank you.
 

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