Question

Two bodies A and B of equal masses m lie on a smooth frictionless horizontal table. They are initially separated by a distance 2l and are connected by a thread of length 2l. The thread is pulled at its midpoint O by a constant force F at right angle to AB. Find the acceleration of A and B in the direction at right angles to the direction of F as a function of the displacement(x) of the midpoint O

Answer 1

Answer on Question #47505-Physics-Mechanics-Kinematics-Dynamics

Two bodies A and B of equal masses $m$ lie on a smooth frictionless horizontal table. They are initially separated by a distance $2l$ and are connected by a thread of length $2l$. The thread is pulled at its midpoint O by a constant force $F$ at right angle to AB. Find the acceleration of A and B in the direction at right angles to the direction of $F$ as a function of the displacement $(x)$ of the midpoint O

Solution

When the thread is pulled the tensions $T$ of the halves of the thread acts on bodies A and B. In y-direction (at right angle to AB):

$$F = 2T \cos \alpha,$$

where $\cos \alpha = \frac{x}{l}$.

In z-direction (at right angles to the direction of $F$):

$$m a_z = T \sin \alpha.$$

The acceleration of A and B in the direction at right angles to the direction of $F$ is

$$a_z = \frac{T}{m} \sin \alpha = \frac{F}{2m} \frac{\sin \alpha}{\cos \alpha} = \frac{F}{2m} \frac{\sqrt{1 - \left(\frac{x}{l}\right)^2}}{\frac{x}{l}} = \frac{F}{2m} \sqrt{\left(\frac{l}{x}\right)^2 - 1}, \text{ when } x \leq l \text{ and } 0 \text{ when } x > l.$$

Answer: $\left\{ \begin{array}{ll} \frac{F}{2m} \sqrt{\left(\frac{l}{x}\right)^2 - 1}, & x \leq l \\ 0, & x > l. \end{array} \right.$

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