Question

A light smooth inextensible string connected with two identical particles each or mass m passes through a light ring R Ifa force F
pulls ring the relative acceleration between the particles has magnitude
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Answer 1

Answer:

The relative acceleration between the particles has magnitude ∣$\frac{Ftan\theta}{m}$∣

Explanation:

Tension force acting on both string and taking component of t writing  equation of both mass and mass less string (assumed).

F = 2T cosθ

$\bold {T = \frac{F}{2cos\theta}}$

• Acceleration on upper mass m,

$\bold {{a_u = a_x + a_y} = \frac{F_x}{m} - \frac{F_y}{m} = \frac{Tcos\theta}{m} - \frac{Tsin\theta}{m}}$  ⇒ -ve sign indicates -Y axis.

• Acceleration on lower mass m,

$\bold {{a_L = a_x + a_y} = \frac{F_x}{m} - \frac{F_y}{m} = \frac{Tcos\theta}{m} + \frac{Tsin\theta}{m}}$ ⇒ +ve sign indicates +Y axis.

∣$a_u -a_L$∣ = ∣ $\frac{Tcos\theta}{m}$ - $\frac{Tsin\theta}{m}$ - $\frac{Tcos\theta}{m}$ - $\frac{Tsin\theta}{m}$∣

= ∣ - $\frac{Tsin\theta}{m}$ - $\frac{Tsin\theta}{m}$ ∣

= ∣ $\frac{2Tsin\theta}{m}$∣

= ∣$\frac{Ftan\theta}{m}$∣

Answer 2

Answer:

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