Physics Help Forum Acceleration of Blocks on an inclined plane

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 Oct 30th 2017, 05:46 PM #1 Junior Member   Join Date: Oct 2017 Posts: 8 Acceleration of Blocks on an inclined plane Problem 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 Attempt at a solution $\displaystyle m1-x:T2on1-m1sinΘ=m1a1$ $\displaystyle m1-y:n2-m1gcosΘ=0$ $\displaystyle m2-x:T1on2-m2gcosΘ=m2a2$ $\displaystyle m2-y:ns-n1-m2g=0$ 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.
 Nov 1st 2017, 06:53 AM #2 Senior Member     Join Date: Aug 2008 Posts: 113 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. $m_1 g \sin{\theta} - T = m_1 a$ $T - m_2 g \sin{\theta} = m_2 a$ —————————————————— $g\sin{\theta}(m_1-m_2) = a(m_1+m_2)$ $a=\dfrac{g\sin{\theta}(m_1-m_2)}{m_1+m_2}$ finish by substituting the above expression for acceleration into either of the two initial equations and solve for tension, $T$. topsquark likes this.

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