# Accleration of 2blocks

1. Oct 12, 2011

### ckboii89

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

2. Relevant equations
a= F/m

F= mgsinθ-Ffr

Ffr = $\mu$mgcosθ

d= 1/2gt^2

3. The attempt at a solution

a = (m1gsinθ+m2gsinθ)-($\mu$m1gcosθ+$\mu$m2gcosθ)/(m1+m2)

a = 3.72 m/s^2

t = 1.03s

I need someone to verify if this is correct, no anwser key was provided

d=

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2. Oct 12, 2011

### ehild

Take care with the formulation. The first term has to be divided by (m1+m2), too. Your result is not correct.

ehild

Last edited: Oct 12, 2011
3. Oct 12, 2011

### ckboii89

a=[gsinθ-(μ1m1gcosθ+μm2gcosθ)]
a=.298m/s^2

t=3.66

4. Oct 12, 2011

### ehild

Still wrong. Did you divide the second term with m1+m2?

ehild

5. Oct 12, 2011

### ckboii89

a=[gsinθ-g(μ1cosθ+μcosθ)]
a = 2.35 m/s^2

t = 1.30s

ck

6. Oct 12, 2011

### ehild

Correct the red part.

ehild

7. Oct 12, 2011

### ckboii89

[gsinθ-μ1gcosθ-μ2gcosθ]
?

ehild i really appreciate the help

is the contact force the difference between the forces of the blocks? or the sum

8. Oct 12, 2011

### ehild

The contact force is the magnitude of the force one block exerts to the other.

The common acceleration is the sum of all forces on the system divided by the sum of the masses. The internal forces (the force of block 1 on block 2 and the one block2 exerts to block 1 are opposite and of equal magnitude) cancel, so a=∑Fi(external)]/mi.

If T is the contact force, Newton's second law applied to both blocks says that

m1a1=m1gsin(θ)-T-m1μ1gcos(θ) and

m2a1=m2gsin(θ)+T-m2μ2gcos(θ) .

If the blocks move together, a1=a2.
$$a=gsin(\theta)-\frac{m_1 \mu_1+m_2\mu_2}{m_1 +m_2}g cos(\theta)$$