# Help me please, Two particles each of mass m, are joined by a thin string of length 2L, as shown in Figure 48. A uniform force F is applied in the middle of the string (x = 0) forming a right angle with the initial position of the string. Show that th????

## Two particles each of mass m, are joined by a thin string of length 2L, as shown in Figure 48. A uniform force F is applied in the middle of the string (x = 0) forming a right angle with the initial position of the string. Show that the acceleration of each mass in the direction of 90 degrees with F is given ${a}_{x} = \frac{F}{2 m} \frac{x}{{L}^{2} - {x}^{2}} ^ \left(\frac{1}{2}\right)$ May 29, 2018

See below

#### Explanation: At all times, where co-ordinate $y$ points in the same direction as $F$, Newton's Law for the entire system is:

• $F = 2 m \ddot{y} \implies \ddot{y} = \frac{F}{2 m}$

For each particle, resolving the tension $T$ in the string in the direction of, and perpendicular to, motion:

• $T \cos \alpha = m \ddot{y}$

• $T \sin \alpha = - m \ddot{x}$

$\implies \ddot{x} = - \ddot{y} \tan \alpha = - \frac{F}{2 m} \tan \alpha$

From the annotated drawing:

• $\tan \alpha = \frac{x}{\sqrt{{L}^{2} - {x}^{2}}}$

$\implies \ddot{x} = - \frac{F}{2 m} \frac{x}{\sqrt{{L}^{2} - {x}^{2}}} q \quad \square$

(NB: There is a minus sign as the $x$ direction points left to right away from the axis of symmetry at $x = 0$. The accelerating force is $- T \sin \alpha \setminus \boldsymbol{\hat{x}}$ so that must be correct.)

The Spanish bit then asks about $x = L$, which is right at the start of the motion

Well:

${\lim}_{x \to L} \left\{\begin{matrix}\ddot{x} = \infty \\ T = \infty \\ \ddot{y} = 0\end{matrix}\right.$

The solution appear to blow up.

But this equation doesn't mean anything at that point in time, as it is derived from there being some angle $\alpha \ne 0$, etc