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Old Oct 30th 2017, 06:46 PM   #1
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Acceleration of Blocks on an inclined plane

The figure below shows a block of mass m1 sliding on a block of mass m2. The inclined surface is fixed, and the angle of the incline is θ. All surfaces are frictionless Find the acceleration of each block and the tension in the
string that connects the blocks
Acceleration of Blocks on an inclined plane-sliding-blocks.png

Attempt at a solution
Acceleration of Blocks on an inclined plane-sliding-block-diagrams.png
$\displaystyle m1-x:T2on1-m1sinΘ=m1a1$
$\displaystyle m1-y:n2-m1gcosΘ=0$
$\displaystyle m2-x:T1on2-m2gcosΘ=m2a2$
$\displaystyle m2-y:ns-n1-m2g=0$
Newton's III:$\displaystyle T1on2=T2on1$
Acceleration constraint:$\displaystyle a2=-a1$

$\displaystyle =>T1on2=m2gcosΘ-m2a1$
$\displaystyle =>m2gcosΘ-m1sinΘ=m2a1+m1a1$
$\displaystyle =>m2gcosΘ-m1sinΘ=a1(m2+m1)$
$\displaystyle =>(m2gcosΘ-m1sinΘ)/(m2+m1)=a1$

So since a2=-a1, would a2 just be the negative of the expression I found above? Also I'm not really sure how to find the tension from what is given.
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Old Nov 1st 2017, 07:53 AM   #2
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Looks like a simple Atwood machine oriented at an angle instead of its commonly viewed vertical placement.

Assuming the pulley and string have negligible mass and $m_1 > m_2$, the following system of scalar equations can be used to determine the magnitudes of acceleration and tension which are the same for both masses.

$m_1 g \sin{\theta} - T = m_1 a$
$T - m_2 g \sin{\theta} = m_2 a$
$g\sin{\theta}(m_1-m_2) = a(m_1+m_2)$


finish by substituting the above expression for acceleration into either of the two initial equations and solve for tension, $T$.
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acceleration, blocks, inclined, plane

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