Stuck on algebra to N2L solution

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The discussion revolves around solving a physics problem from Halliday's textbook involving Newton's Second Law (N2L) equations. The user is struggling to connect four given N2L equations to derive the friction force equation. A hint is provided to solve the first and third equations for acceleration, then set them equal to find the friction force. The solution involves substituting the derived expressions back into the equations to simplify and solve for the desired variable. The user expresses gratitude for the clarification, indicating a better understanding of the problem-solving process.
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I already understand the problem, but am having trouble with the algebra for textbook Fund of Physics Solutions, Halliday, 8e, Ch6 Prob 34. Copied here

http://i56.tinypic.com/20uwhw4.jpg

I've been staring at this solution to N2L for hours and can't figure out how to go from the top 4 N2L equations to the bottom equation for f (friction). I tried making both 'as' and 'ab' as 'a' but keep geting back to the orig equation.

Just give me a hint, please...
 
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o.k., i typed it here, but it's cryptic with (subscripts)

these four N2L equations:

-f = m(s) a(s)
F(ns) -F(Nb) -m(s)g = 0
f - F = m(b) a(b)
F(Nb) - m(b) g

are shown in the soln to resolve to

f = ( m(s) F ) / ( (m(s) + m(b) )

where do i begin ?
 
How are the slab and block related?
 
oh ! now i see. it's checking if the block on top is sliding on the slab. not deriving it.
 
NamaeKana said:
o.k., i typed it here, but it's cryptic with (subscripts)

these four N2L equations:

-f = m(s) a(s)
F(ns) -F(Nb) -m(s)g = 0
f - F = m(b) a(b)
F(Nb) - m(b) g

are shown in the sol'n to resolve to

f = ( m(s) F ) / ( (m(s) + m(b) )

where do i begin ?

What have you tried?

The solution says that as = ab, they drop the subscripts & simply call them both "a".

Solve the 1st equation for a: a=\frac{-f}{m_s}\ .

Solve the 3rd equation for a: a=\frac{f-F}{m_b}\ .

Set the right hand sides equal to each other & solve for f .

Added in Edit:

It's probably easier Algebra-wise to do the following.

Solve the 1st equation for f:  f = -ms·a

Plug this into the 3rd equation:  -ms·a -F = mb·a

Solve for a, then plug that back into the equation: f = -ms·a
 
Last edited:
thanks. i wouldn't have figured that out by myself in a millions years ;-). i get the logic, solve 1 & 3 for a, then take a and plug back into 1.
 

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