Derive an expression for the magnitude of the acceleration of an Atwood\'s machi

Question

Derive an expression for the magnitude of the acceleration of an Atwood's machine. Assume that m1>m2. Put the answer in terms of g, delta m=(m1-m2) and m total =(m1+m2)

Explanation / Answer

Equation for constant acceleration:- The free body diagrams of the two hanging masses of the Atwood machine. Our sign convention, depicted by the acceleration vectors is that m1 accelerates downward and that m2 accelerates upward, as would be the case if m1 > m2 We are able to derive an equation for the acceleration by using force analysis. If we consider a massless, inextensible string and an ideal massless pulley, the only forces we have to consider are: tension force (T), and the weight of the two masses (W1 and W2). To find an acceleration we need to consider the forces affecting each individual mass. Using Newton's second law (with a sign convention of m1 > m2) we can derive a system of equations for the acceleration (a). As a sign convention, we assume that a is positive when downward for m1, and that a is positive when upward for m2. Weight of m1 and m2 is simply W1 = m1g and W2 = m2g respectively. Forces affecting m1: m1g-T=m1a Forces affecting m2: T-m2g=m2a and adding the two previous equations we obtain m1g-m2g=m1a+m2a, and our concluding formula for acceleration a = g{(m1-m2)/(m1+m2)} Conversely, the acceleration due to gravity, g, can be found by timing the movement of the weights, and calculating a value for the uniform acceleration a: d = {1/2} at^2 . Equation for tension:- It can be useful to know an equation for the tension in the string. To evaluate tension we substitute the equation for acceleration in either of the 2 force equations. a = g{(m1-m2)/(m1 + m2)} For example substituting into m1a = T ? m1g, we get T={(2gm1m2)/(m1 + m2)} The tension can be found in using this method.

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