Two masses $m_{1}$ and $m_{2}$ are connected with a string passing over a frictionless pully fixed at the corner of the table as shown in the figure, The coefficient of static friction of mass $m_{1}$ with the table is $μ s$ . Calculate the minimum mass $m_{3}$ that may be placed on $m_{1}$ to prevent it from sliding. Check if $m_{1} = 15$ kg, $m_{2} = 10$ kg, $m_{3} = 25$ and $μ_{s} = 0.2$ .

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Solution

$T = m_{2} g$ for no motion

At $m_{1}$ , normal force $N = (m_{1} + m_{3}) g$

friction $= f \leq μ_{s} N = μ_{s} (m_{1} + m_{3}) g$

For no sliding-

$f_{1} = T$

$⟹ μ_{s} (m_{1} + m_{3}) \geq m_{2} g$

$⟹ m_{3} \geq \frac{m_{1}}{μ_{s}} - m_{1}$

$⟹ m_{3} \geq \frac{10}{0.2} - 15$

$⟹ m_{3} \geq 35 k g$ This is minimum value to prevent sliding.

For $m_{3} = 25 k g < 35 k g$ , blocks will slide

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