Forces/Dynamics: A block sliding on a plate

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Homework Help Overview

The discussion revolves around a physics problem involving forces and dynamics, specifically a block sliding on a plate. The scenario includes a block of mass 0.5 kg on a frictionless plate with an external force applied over time, leading to questions about static and kinetic friction, acceleration, and distance traveled after the block begins to slide.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the forces acting on the block and plate, questioning how acceleration affects the system. They discuss the relationship between external force, static friction, and the resulting accelerations of both the block and the plate.

Discussion Status

Participants are actively engaging with the problem, raising questions about the calculations for forces and accelerations. Some have reached tentative conclusions about specific values, while others seek clarification on the implications of the applied force and the conditions under which the block begins to slide.

Contextual Notes

There is an ongoing discussion about the initial conditions of the system, particularly regarding the velocities and accelerations of the block and plate once the block begins to slide. Participants are also considering the implications of the frictionless surface beneath the plate and how it affects the overall dynamics.

  • #31
haruspex said:
No.
In post #5 you had two equations which I will abbreviate as ##a=\frac{F-f_s}{m_b}## and ##a=\frac{f_s}{m_p}##.
Previously, you combined them so as to eliminate a and found the relationship between F and fs. Now I am asking you to combine them so as to eliminate fs instead, and hence find the relationship between F and a.
Ok.. If I understand your statement correctly, do you mean by:
##a=\frac{F-f_s}{m_b}## and ##a=\frac{f_s}{m_p}##
##m_b*a_b= F-f_s##
##f_s = F - (m_b*a_b)##
##a_p = \frac {(F - (m_b*a_b))}{m_p}##
?
Thank you so much for your patience and guidance :smile:
 
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  • #32
jisbon said:
Ok.. If I understand your statement correctly, do you mean by:
##a=\frac{F-f_s}{m_b}## and ##a=\frac{f_s}{m_p}##
##m_b*a_b= F-f_s##
##f_s = F - (m_b*a_b)##
##a_p = \frac {(F - (m_b*a_b))}{m_p}##
?
Thank you so much for your patience and guidance :smile:
Yes, but remember we are dealing with the time up to when slipping starts, so the two accelerations are the same.
Get it into the form a=..., and substitute the given formula for F as a function of t.
 
  • #33
haruspex said:
Yes, but remember we are dealing with the time up to when slipping starts, so the two accelerations are the same.
Get it into the form a=..., and substitute the given formula for F as a function of t.
##m_b*a_b= F-f_s##
##f_s = F - (m_b*a_b)##
##a_p = \frac {(F - (m_b*a_b))}{m_p}##
Referring to your highlighted part, if I wish to substitute ##F = 2tN##
Should't I just take ##f_s = F - (m_b*a_b)## = ##F= f_s + (m_b*a_b)## = ##2tN##?
I feel that my though processes are just different from yours. So I would like to check the process again:

1) I need to find distance traveled by plate for 1s after it slips.
2) So I need to find out the velocity of the plate at the moment the block slips, and the acceleration of the plate at that point (which will be constant since only friction is involved for plate)
3)I can find the acceleration of the plate by using, F=ma, thus acceleration of plate is #5.88N/1.5## . Is this correct? (I'm thinking that acceleration is ALWAYS constant for plate since it only experience friction all the time)
4) Then to find out velocity of plate, I need to find out time it takes for block to slip. In this case, is finding t from:
##F= f_s + (m_b*a_b)## = ##2tN## correct?
If I can find time taken for block to slip, won't the velocity of plate when block slip = ##v=u+at## where a is found in step 3?

Thank you.
 
  • #34
jisbon said:
##m_b*a_b= F-f_s##
##f_s = F - (m_b*a_b)##
##a_p = \frac {(F - (m_b*a_b))}{m_p}##
You seem to have missed the part where I pointed out that until the block slips the two accelerations are the same. There is not an ap and an ab, just a. So get it into the form a=(a function of masses and F) then use F=2t.
 
  • #35
haruspex said:
You seem to have missed the part where I pointed out that until the block slips the two accelerations are the same. There is not an ap and an ab, just a. So get it into the form a=(a function of masses and F) then use F=2t.
So something like:
##a = \frac {(F - (m_b*a_b))}{m_p}##
##a = \frac {(2TN - (m_b*a_b))}{m_p}##
?
 
  • #36
jisbon said:
So something like:
##a = \frac {(F - (m_b*a_b))}{m_p}##
##a = \frac {(2TN - (m_b*a_b))}{m_p}##
?
Why do you still have ab terms on the right? There is only the one acceleration during this phase of the problem. Turn all ap and ab terms into plain a and rearrange to get the equation into the form a=some function of F, mb and mp. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.
 
  • #37
haruspex said:
Why do you still have ab terms on the right? There is only the one acceleration during this phase of the problem. Turn all ap and ab terms into plain a and rearrange to get the equation into the form a=some function of F, mb and mp. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.
Oh sorry. Corrected it as followed:
##a = \frac {(F - (m_b*a))}{m_p}##
##a*m_p = (F - (m_b*a))##
##(a*m_p) + (m_b*a) = F##
##a(m_p + m_b) = F##
##a = \frac {F}{(m_p + m_b)}##
##a = \frac {2tN}{(m_p + m_b)}##
To be clear, I will be using N for the plate right?
This a I'm finding will be the acceleration before it slips. So from there, I am able to find the velocity of plate when it starts to slip with this acceleration value? Thanks!
 
  • #38
jisbon said:
from there, I am able to find the velocity of plate when it starts to slip with this acceleration value?
I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?
 
  • #39
haruspex said:
I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?
I will integrate it to become velocity?
So from my previous equation:
jisbon said:
##a=\frac {2tN}{m_p+m_b}##
##a = \frac {2t*9.8*1.5}{1.5+0.5}##
##a = 14.7t##
##v = (\frac {14.7}{2})t^2##?
 
  • #40
jisbon said:
I will integrate it to become velocity?
So from my previous equation:

##a = \frac {2t*9.8*1.5}{1.5+0.5}##
##a = 14.7t##
##v = (\frac {14.7}{2})t^2##?
Yes, your integration is fine, but where did the factor g come from?
 
  • #41
haruspex said:
Yes, your integration is fine, but where did the factor g come from?
Oh wait isn't ##N= g* mass##?
 
  • #42
jisbon said:
Oh wait isn't ##N= g* mass##?
No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s2. Nothing to do with gravity.
 
  • #43
haruspex said:
No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s2. Nothing to do with gravity.
Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will ##a= \frac {2t}{1.5+0.5} = t##
Hence ##v = \frac {1}{2}t^2##
?
 
  • #44
jisbon said:
Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will ##a= \frac {2t}{1.5+0.5} = t##
Hence ##v = \frac {1}{2}t^2##
?
Yes,
 
  • #45
So with a and v, I can find out initial velocity when plate before it slips?
 
  • #46
jisbon said:
So with a and v, I can find out initial velocity when plate before it slips?
Yes - do you see how?
 

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