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(A) For pure translational motion,the force must be applied at the center of mass of the body.Let the mass of the horizontal rod $AB$ be $m$. Since th

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$\frac{4}{3} \ell$

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(A) For pure translational motion,the force must be applied at the center of mass of the body.Let the mass of the horizontal rod $AB$ be $m$. Since the vertical rod $CD$ has length $2\ell$ and the horizontal rod $AB$ has length $\ell$,assuming uniform density,the mass of rod $CD$ will be $2m$.Let $y_1$ be the center of mass of rod $AB$ and $y_2$ be the center of mass of rod $CD$.Taking point $C$ as the origin $(0,0)$:The center of mass of rod $AB$ is at a distance of $2\ell$ from $C$ along the vertical axis,so $y_1 = 2\ell$.The center of mass of rod $CD$ is at a distance of $\ell$ from $C$ along the vertical axis,so $y_2 = \ell$.The position of the center of mass of the system along the vertical axis is given by:$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$Here,$m_1 = m$ (for rod $AB$) and $m_2 = 2m$ (for rod $CD$).$y_{cm} = \frac{m(2\ell) + (2m)(\ell)}{m + 2m} = \frac{2m\ell + 2m\ell}{3m} = \frac{4m\ell}{3m} = \frac{4}{3}\ell$.Thus,the force must be applied at a distance of $\frac{4}{3}\ell$ from $C$.

$A$ $T$-shaped object with dimensions shown in the figure is lying on a smooth floor. $A$ force $\vec{F}$ is applied at the 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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