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Am i wrong or is it the book? newtonian physics two stacked masses

  • Thread starter clark1089
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Homework Statement


A 2.0 kg block rest on a 4.0kg block that is on a frictionless horizontal table. The coefficients of friction are u_s = 0.3 and u_k = 0.2. What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slip? If F has half this value, find the acceleration of each block and the force of friction acting on each block .If F is twice the value you determined in the first question, find the acceleration of each block.


Homework Equations


Fx = F = m1a
Fx = ff = m2a


The Attempt at a Solution


What I need help with is the final part of this problem... part C. Here's what is needed to solve:
I solved for 18 N in part(a), the first part, as the maximum force required to move the object.
Now, I solved for the acceleration of block 2 in the last part; block 2 = ff/m2 = u_k*g = (.2)(9.81) = 2.0 m/s^2

But now I'm trying to solve for the acceleration of the 4.0kg block
F - ff = m1a1 + m2a2
(F - ff - m2a2)/m1 = a1 = 7.019 m/s^2
Not right... the book says it is 7.9 m/s^2. So I tried again ignoring ff
(F - m2a2)/m1 = a1 = 8.019 m/s^2

where am i going wrong here? sorry, no diagram given in problem so I can't scan one.. :X:X

i note that, although they list 18 N as the force, if you leave it unrounded:
17.76 *2 = 35.52
35.52 - m2 = m1a1
35.52 - 4 = 4a
a = 7.88 = 7.9 m/s^2
that can't be it, can it?
 
Last edited:

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
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Homework Statement


A 2.0 kg block rest on a 4.0kg block that is on a frictionless horizontal table. The coefficients of friction are u_s = 0.3 and u_k = 0.2. What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slip? If F has half this value, find the acceleration of each block and the force of friction acting on each block .If F is twice the value you determined in the first question, find the acceleration of each block.


Homework Equations


Fx = F = m1a
Fx = ff = m2a


The Attempt at a Solution


What I need help with is the final part of this problem... part C. Here's what is needed to solve:
I solved for 18 N in part(a), the first part, as the maximum force required to move the object.
Now, I solved for the acceleration of block 2 in the last part; block 2 = ff/m2 = u_k*g = (.2)(9.81) = 2.0 m/s^2

But now I'm trying to solve for the acceleration of the 4.0kg block
F - ff = m1a1 + m2a2
(F - ff - m2a2)/m1 = a1 = 7.019 m/s^2
Not right... the book says it is 7.9 m/s^2. So I tried again ignoring ff
(F - m2a2)/m1 = a1 = 8.019 m/s^2

where am i going wrong here? sorry, no diagram given in problem so I can't scan one.. :X:X

i note that, although they list 18 N as the force, if you leave it unrounded:
17.76 *2 = 35.52
35.52 - m2 = m1a1
35.52 - 4 = 4a
a = 7.88 = 7.9 m/s^2
that can't be it, can it?
You are not drawing Free Body Diagrams (FBD), which are essential to the solution. For the last part, draw a FBD of the bottom block (isolate that block), and look at the forces in the x direction acting on that block. The applied Force F acts right, and the friction force ff acts left. The mass of the block is m1. Solve for a1 using Newton 2. Note that in this diagram and equation for the bottom block, m2 does not come into play.
 
  • #3
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...wow! thank you! :D i got the right answer :)
 

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