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(A) For an object to undergo pure translational motion,the applied force must pass through its center of mass $(CM)$.Let the mass of the horizontal ro

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$\frac{4l}{3}$

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(A) For an object to undergo pure translational motion,the applied force must pass through its center of mass $(CM)$.Let the mass of the horizontal rod $AB$ be $m$. Since the vertical rod $CD$ has length $2l$ and the horizontal rod $AB$ has length $l$,assuming uniform density,the mass of rod $CD$ is $2m$.The center of mass of rod $AB$ is at its midpoint $D$,which is at a distance $2l$ from $C$. Let this be $y_1 = 2l$.The center of mass of rod $CD$ is at its midpoint,which is at a distance $l$ from $C$. Let this be $y_2 = l$.Taking $C$ as the origin,the position of the center of mass of the system along the vertical axis is:$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$$y_{cm} = \frac{m(2l) + (2m)(l)}{m + 2m} = \frac{2ml + 2ml}{3m} = \frac{4ml}{3m} = \frac{4l}{3}$Thus,the force must be applied at a distance of $\frac{4l}{3}$ from $C$ to ensure pure translational motion.

$A$ $T$-shaped object with dimensions shown in the figure is lying on a smooth floor. $A$ force $\vec{F}$ is applied at point $P$ parallel to $AB$,such that the object has only translational motion without rotation. Find the location of $P$ with respect to $C$.

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