![](\"resources/question-resources/pulley1.png\")
We can draw a diagram to show all the forces acting on each mass and the pulley, and the acceleration. Here, the mass of particle $A$ is $m_1 \\mathrm{kg} = \\var{A} \\mathrm{kg}$ and the mass of particle $m_2 \\mathrm{kg} = \\var{B} \\mathrm{kg}$. As particle $B$ is heavier the acceleration will act in the direction shown. The tension in the string is $T \\mathrm{N}$ and the force exerted on the string by the pulley is $F \\mathrm{N}$.
\n
a) We can not treat the whole system as a particle because the particles are moving in different directions. Therefore to find the acceleration we need to resolve for each mass separately.
\nFor $A$ we have equation (1): \\begin{align} T - m_1g & = m_1 a, \\\\
T - \\var{A}g & = \\var{A}a. \\end{align}
For $B$ we have equation (2): \\begin{align} m_2g - T & = m_2a, \\\\
\\var{B}g - T & = \\var{B}a. \\end{align}
In both (1) and (2) we have resolved in the direction of acceleration.
\nWe can solve these two equations simultaneously. For example, by adding (1) and (2) we get
\n\\begin{align} T - \\var{A}g + \\var{B}g - T & = (\\var{A} + \\var{B}) a, \\\\
(\\var{B} - \\var{A})g & = (\\var{A} + \\var{B}) a, \\\\
a & = \\frac{\\var{A} + \\var{B}}{(\\var{B} - \\var{A})g}, \\\\
& = \\var{precround((B*g-A*g)/(A+B),3)} \\mathrm{ms^{-2}}. \\end{align}
The acceleration of each mass is $\\var{precround((B*g-A*g)/(A+B),3)} \\mathrm{ms^{-2}}$.
\nb) To find the tension in the string we can use our answer to part a) in either equation (1) or (2).
\nFrom equation (1) we have
\n\\begin{align} T - \\var{A}g & = \\var{A}a, \\\\
T & = \\var{A} \\times (g + a), \\\\
& = \\var{A} \\times (9.8 + \\var{ac}), \\\\
& = \\var{precround(A*(9.8+ac),3)} \\mathrm{N}. \\end{align}
The tension in the string is $\\var{T}\\mathrm{N}$.
\nc) The force exerted on the pulley by the string is $2T \\mathrm{N} = \\var{precround(2*T,3)} \\mathrm{N}$.
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\nIn the following answer to 3d.p. and take the acceleration due to gravity as $g = 9.8 \\mathrm{ms^{-2}}$.
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