Suppose we had chosen another point through which the rotation axis were to pass. For example, suppose the axis is perpendicular to the page and passes through the location of the father. Does this change the results to parts (A) and (B)?

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Part (A) is unaffected because the calculation of the net force does not involve a rotation axis. In part (B), we would conceptually expect there to be no change if a different rotation axis is chosen because the second condition of equilibrium claims that the torque is zero about any rotation axis.
Let us verify this mathematically. Recall that the sign of the torque associated with a force is positive if that force tends to rotate the system counterclockwise, while the sign of the torque is negative if the force tends to rotate the system clockwise. In the case of a rotation axis passing through the location of the father,[latex] \Sigma au=0 [/latex] yields

[latex] n(d)-(M g)(d)-\left(m_{d} g ight)(d+\ell / 2)=0 [/latex].

[latex] ext { From part (A) we know that } n=m_{f} g+m_{d} g+M g ext {. Thus, we } [/latex] can substitute this expression for n and solve for d:

[latex] \left(m_{f} g+m_{d} g+M g ight)(d)-(M g)(d)-\left(m_{d g} ight)\left(d+\frac{\ell}{2} ight)=0 [/latex].

[latex] \left(m_{f} g ight)(d)-\left(m_{d g} g ight)\left(\frac{\ell}{2} ight)=0 [/latex].

[latex] d=\left(\frac{m_{d}}{m_{f}} ight) \frac{1}{2} \ell [/latex].

This result is in agreement with the one we obtained in part (B).

Question 12.WI.1: Suppose we had chosen another point through which the rotati...

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