How to Solve Dynamics Exercise Involving Force Representation?

[srnixo]

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).​

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

Then, i found F⁰ = 5N (question 1)

After that, (question 2) + (question 3)

I hope it's even correct.

 

[erobz]

Would it be possible for you to concisely give the answers you get for each question you are in doubt about?

 

[srnixo]

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.

 

[erobz]

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.

 

[srnixo]

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 )

 

[erobz]

Before block 2 moves the spring compresses by amount $x_1$

 

[srnixo]

Before block 2 moves the spring compresses by amount $x_1$

Yep which is Force T = Tension, i know this.

 

[erobz]

Yep which is Force T = Tension, i know this.

Ok, then write Newtons 2nd for block 1.

 

[srnixo]

Ok, then write Newtons 2nd for block 1.

And then!!!

 

[erobz]

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...

 

[srnixo]

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.

 

[erobz]

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...

 

[srnixo]

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 :(

 

[erobz]

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.

 

[srnixo]

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

 

[erobz]

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.

 

[srnixo]

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

 

[erobz]

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 $\mu_s, k, M, g$

 

[srnixo]

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?

 

[erobz]

Isn't 4N huh?

No, $T$ is not $4 \rm{N}$.

$x_1$?

 

[srnixo]

No, $T$ is not $4 \rm{N}$.

$x_1$?

Then it's 5N , i considered the F⁰ case.

 

[erobz]

Then it's 5N , i considered the F⁰ case.

Yes, its static friction acting on block 2...it's not moving.

 

[srnixo]

Yep yep what about question 5 mr. Do u know anything about it! I actually didn't understand anything about it

 

[erobz]

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 $\Delta x$.

 

[srnixo]

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

 

[erobz]

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.

 

[srnixo]

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.

 

[kuruman]

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.

 

[haruspex]

First, a nitpick: you keep writing "tension", but the spring is under compression.

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.

 

[erobz]

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.