Two weights $2 \mathrm{~N}$ and $3 \mathrm{~N}$ are suspended from the ends of an inextensible string…
Two weights $2 \mathrm{~N}$ and $3 \mathrm{~N}$ are suspended from the ends of an inextensible string passing over a fixed frictionless pulley. If the pulley is pulled up with an acceleration equal to the acceleration due to gravity, then the tension in the string is
$2.4 \mathrm{~N}$
$5.0 \mathrm{~N}$
$4.8 \mathrm{~N}$
$6.0 \mathrm{~N}$
Solution
The masses of the two weights are determined from their given weights: \(m_{1}=\frac{W_{1}}{g}=\frac{2\text{\ N}}{g}\) \(m_{2}=\frac{W_{2}}{g}=\frac{3\text{\ N}}{g}\) The effective acceleration of the system is considered in the frame of reference of the pulley. Since the pulley is pulled up with acceleration \(g\), the effective acceleration due to gravity for the masses becomes \(g^{\prime }=g+g=2g\). The formula for tension in an Atwood machine with an upward accelerating pulley is used: \(T=\frac{2m_{1}m_{2}}{m_{1}+m_{2}}(g+a)\) where \(a\) is the acceleration of the pulley, which is \(g\). Substituting the values of \(m_{1}\) and \(m_{2}\): \(T=\frac{2\left(\frac{2}{g}\right)\left(\frac{3}{g}\right)}{\frac{2}{g}+\frac{3}{g}}(g+g)\) The expression is simplified: \(T=\frac{2\left(\frac{6}{g^{2}}\right)}{\frac{5}{g}}(2g)\) \(T=\frac{12}{g^{2}}\times \frac{g}{5}\times 2g\) \(T=\frac{12\times 2}{5}\) \(T=\frac{24}{5}\) \(T=4.8\text{\ N}\)