# Acceleration of the lower block

1. May 15, 2009

### brunettegurl

1. The problem statement, all variables and given/known data

In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 58.6 N. The coefficient of kinetic friction between the lower block and the surface is 0.204. The coefficient of kinetic friction between the lower block and the upper block is also 0.204. What is the acceleration of the lower block, if the mass of the lower block is 4.68 kg and the mass of the upper block is 2.11 kg?

2. Relevant equations

F=ma / friction = $$\mu$$mg

3. The attempt at a solution

so i wrote the forces that affect each box

For the upper box ::$$\sum$$ F= T-$$\mu$$m1g
for the lower box :: $$\sum$$ F= F-T- friction1- friction2

i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what i'm doin wrong thanx

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2. May 15, 2009

### Staff: Mentor

OK, but what did you use for the friction forces on the lower box?
I assume you applied Newton's 2nd law to each box to get two equations. Show exactly what you did and then we can see what went wrong.

3. May 15, 2009

### brunettegurl

for friction 1 i used (coeffecient of frictionxmass1xgravity) and for friction 2 i use (coeffecient of frictionxmass2xgravity)
and the equation i got before isolating for acceleration is ::

a(m2+m1)= Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity)

so then a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)

4. May 15, 2009

### Staff: Mentor

OK.
Careful: What's the normal force between the bottom block and the floor?

5. May 15, 2009

### brunettegurl

is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...???

6. May 15, 2009

### Staff: Mentor

Yes, the normal force is the combined weight of both blocks, thus f2 = μ(m1 + m2)g.

7. May 15, 2009

### brunettegurl

ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??

8. May 15, 2009

### Staff: Mentor

9. May 15, 2009

### brunettegurl

my final equation wld look like ::
m2a = [Fapplied-m1a-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)]

a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)

Fapplied is the force given in the question

10. May 15, 2009

### Staff: Mentor

Your equations are a bit confusing, due to what may be a few typos. (The '=', for one.) Can you rewrite using F, m1, m2, μ, and g? Try to simplify as much as possible.

11. May 15, 2009

### brunettegurl

m2a= [F -m1a- (μm1g)- (μg(m1+m2)) - (μm2g)]

a = [F - (μm1g) - (μg(m1+m2)) - (μm2g)]/(m1+m2)

iss that clearer ??

12. May 15, 2009

### Staff: Mentor

Yes, much clearer. You have an extra m2 term in there for some reason. If you write the original Newton's law equations for m1 and m2, perhaps we can see where it got snuck in.

13. May 15, 2009

### brunettegurl

this is what i have as my equations for m1 and m2

m1:: m1a= T- μm1g
m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g

i dont see the extra m2 in my equations can u be a bit more specific

14. May 16, 2009

### Staff: Mentor

That second friction term in the m2 equation is the problem--you're using the wrong mass.

15. May 16, 2009

### brunettegurl

so this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g wld be this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm1g....
then the final equation wld look like a = [Fapp - (μm1g) - (μg(m1+m2)) - (μm1g)]/(m1+m2)
is that correct??

16. May 16, 2009

### Staff: Mentor

That looks good. But consolidate some of those terms.