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Friction problem involving 2 blocks sliding in 2 directions with 2 frictions

by 2FAST4U8
Tags: force, friction, newtons law
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2FAST4U8
#1
Nov9-08, 09:24 PM
P: 1
1. The problem statement, all variables and given/known data



A 1-kg block is pushed against a 4-kg block on a horizontal surface of coefficient of friction 0.25, as shown in the figure. Determine the minimum force needed to ensure that the 1-kg block does not slip down. Assume that the coefficient of friction at the interface between the block is 0.4. Hint: The two blocks exert equal and opposite forces on each other.

2. Relevant equations

F=(A+B)a
Ag=fs
N=Ba
Ag=[tex]\mu[/tex]Ba

3. The attempt at a solution

In class we have done similiar problems, only with a frictionless horizontal surface. I don't know how to account for the horizontal coefficient of friction of .25 in this problem. Using the above equations and ignoring the horizontal coefficient of friction, I get this:

a=(Ag)/([tex]\mu[/tex]B

F=(((A+B)Ag)/([tex]\mu[/tex]B))

F=(((4+1)(1)(9.8))/((.4)(4)))

F=30.62N

Now how do I account for the horizontal coefficient of friction of .25? Thanks in advance for any help you can provide.
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asleight
#2
Nov9-08, 09:32 PM
P: 154
Quote Quote by 2FAST4U8 View Post
1. The problem statement, all variables and given/known data



A 1-kg block is pushed against a 4-kg block on a horizontal surface of coefficient of friction 0.25, as shown in the figure. Determine the minimum force needed to ensure that the 1-kg block does not slip down. Assume that the coefficient of friction at the interface between the block is 0.4. Hint: The two blocks exert equal and opposite forces on each other.

2. Relevant equations

F=(A+B)a
Ag=fs
N=Ba
Ag=[tex]\mu[/tex]Ba

3. The attempt at a solution

In class we have done similiar problems, only with a frictionless horizontal surface. I don't know how to account for the horizontal coefficient of friction of .25 in this problem. Using the above equations and ignoring the horizontal coefficient of friction, I get this:

a=(Ag)/([tex]\mu[/tex]B

F=(((A+B)Ag)/([tex]\mu[/tex]B))

F=(((4+1)(1)(9.8))/((.4)(4)))

F=30.62N

Now how do I account for the horizontal coefficient of friction of .25? Thanks in advance for any help you can provide.
Well, examining block one, we notice that we want [tex]\sum\vec{F}=m\vec{a}=0=\vec{F}_f+\vec{F}_g\rightarrow\vec{F}_f=-\vec{F}_g[/tex]. From this, we can examine the frictional force, specifically:

[tex]\vec{F}_f=\mu_i\vec{N}\rightarrow m\vec{g}/\mu_i=\vec{N}[/tex].

What next? :)


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