# How to find the acceleration value for each block when one is over another in an incline?

#### Chemist116

The problem is as follows:

The figure from below shows a block of mass $m_2=2\,kg$ and $m_1=1\,kg$. Assuming the surfaces are frictionless find the value of the acceleration of each block.

The alternatives are as follows:

$\begin{array}{ll} 1.&\frac{g}{7}\\ 2.&\frac{g}{5}\\ 3.&\frac{g}{4}\\ 4.&\frac{g}{6}\\ 5.&\frac{g}{8}\\ \end{array}$

How exactly should I make a FBD (free body diagram) in this system?. I'm not sure exactly how to assess the acceleration for both objects.

Since the object which is on top weighs more than the one which is on bottom then it makes sense that the motion for the object on top will more to the left and the one which is on bottom to move to the right.

$T-(m_1+m_2)g\sin 30^{\circ}=(m_1+m_2)(a)$

while for the object on top it would be:

$m_2g\sin 30^{\circ}-T=m_2(-a)$

But I'm not sure if this is the right interpretation for this system:

If I go further on this it would become into:

$(m_1+m_2)g\sin 30^{\circ}+(m_1+m_2)(a)=T$

$m_2g\sin 30^{\circ}+m_2(a)=T$

$m_2g\sin 30^{\circ}+m_2(a)=(m_1+m_2)g\sin 30^{\circ}+(m_1+m_2)(a)$

$m_2a=m_1g\sin30^\circ+m_1a$

$a=\frac{m_1g\sin30^\circ}{m_2-m_1}=\frac{0.5g}{1}=\frac{1}{2}g$

But it doesn't check with any of the alternatives. Where exactly did I missunderstood?. Can someone help me here?.

#### skeeter

Math Team
How exactly should I make a FBD (free body diagram) in this system?
isolate each mass and sketch a FBD for each one separately

$a = \dfrac{g}{6}$

#### Chemist116

isolate each mass and sketch a FBD for each one separately

$a = \dfrac{g}{6}$
Okay. Here it is:

As mentioned in earlier threads I'm still adjusting my abilities to draw these diagrams. But if I get the picture correctly. Then it would be as follows:

For $m_1$:

$T-m_{1}g\sin 30^{\circ}=m_{1}a$

For $m_{2}$:

$T-m_{2}g\sin 30^{\circ}=m_{2}(-a)$

Then combining both expressions:

$T=m_1a+m_1g\sin 30^\circ$

$T=m_{2}g\sin 30^{\circ}-m_{2}a$

$m_1a+m_1g\sin 30^\circ = m_{2}g\sin 30^{\circ}-m_{2}a$

$a=\frac{m_{2}g\sin 30^\circ-m_1g\sin30^\circ}{m_1+m_2}$

$a=\frac{2\frac{1}{2}g-1\frac{1}{2}g}{3}=\frac{g}{6}$

For which I believe it is what you got. But I'm still confused on why should masses be isolated?. Shouldn't be a contribution of the weight from the object which is on the top affect the one which is on the bottom?. This is the part where I'm confused. Can you explain this better please?. Is my free body diagram correct?.

#### skeeter

Math Team
Shouldn't be a contribution of the weight from the object which is on the top affect the one which is on the bottom?
Only if friction is present.

Is my free body diagram correct?
it's ok, except for the absence of normal forces ... which would come into play if friction were a factor. You also have cosine and sine switched, which made your last equation incorrect.

$\dfrac{2\frac{1}{2}g - 1\frac{1}{2}g}{3} = \dfrac{g}{3}$

Some advice ... setting up equations using magnitudes (scalars) keeps the "negative sign" mistakes to a minimum.

For the top mass ...
(winning force) - (losing force) = net force ... this keeps the terms positive, easier to deal with.
$m_2g \sin(30) - T = m_2a$
it's clear the acceleration will be down the incline, $a$ is the magnitude of that acceleration.

For the bottom mass ...
(winning force) - (losing force) = net force
$T - m_1g \sin(30) - T = m_1a$
it's clear the acceleration will be up the incline, the $a$ is the magnitude of that acceleration.

summation of the two equations ...

$m_2g \sin(30) - T = m_2a$
$T - m_1g \sin(30) - T = m_1a$
---------------------------------------------------------
$(m_2-m_1)g\sin(30) = (m_2+m_1)a$

$(2-1)g \cdot \dfrac{1}{2} = (2+1)a$

$\dfrac{g}{2} = 3a$

$\dfrac{g}{6} = a$