Question

A 'T' shaped object with dimensions shown in the figure, is lying on a smooth floor. A force $\overrightarrow{F}$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

• A. $\frac{3}{2}\ell$
• B. $\frac{2}{3}\ell$
• C. $\ell$
• D. $\frac{4}{3}\ell$

Answer 1

Answer: (C)

$\ell$

Answer 2

For an object to undergo pure translational motion without rotation, the applied force must act at its center of mass. Let's assume the mass of the horizontal bar AB is $m$ and the mass of the vertical bar is $2m$. Let C be the origin (0,0). The vertical bar extends from C to a point at a height of $2\ell$. Its center of mass is at $(0, \ell)$. The horizontal bar AB has length $\ell$ and is attached at the midpoint of the vertical bar, which is at $(0, \ell)$. Assuming it's symmetric, it extends from $(-\ell/2, \ell)$ to $(\ell/2, \ell)$. Its center of mass is at $(0, \ell)$.

The center of mass of the entire T-shaped object is calculated as follows: Mass of horizontal bar ($m_1$) = $m$, Center of Mass ($x_1, y_1$) = $(0, \ell)$ Mass of vertical bar ($m_2$) = $2m$, Center of Mass ($x_2, y_2$) = $(0, \ell)$ Total mass ($M$) = $m_1 + m_2 = m + 2m = 3m$.

$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{M} = \frac{m(0) + 2m(0)}{3m} = 0$ $y_{CM} = \frac{m_1 y_1 + m_2 y_2}{M} = \frac{m(\ell) + 2m(\ell)}{3m} = \frac{3m\ell}{3m} = \ell$

The center of mass of the T-shaped object is at $(0, \ell)$. Since the force $\overrightarrow{F}$ must be applied at the center of mass for pure translation, point P is located at $(0, \ell)$. The location of P with respect to C (the origin) is the distance along the vertical axis, which is $\ell$.