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प्रश्न
Two masses m1 and m2 are connected with a string passing over a frictionless pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m1 with the table is µs Calculate the minimum mass m3 that may be placed on m1 to prevent it from sliding. Check if m1 = 15 kg, m2 = 10 kg, m3 = 25 and µs = 0.2.
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उत्तर
Let m3 is the mass added on m1
Maximal static friction
`f_s^max` = µsN = µs (m1 + m3)g
Here,
N = (m1 + m3)g
Tension acting on string = T = m2 g
Equate (1) and (2)
µs(m1 + m3) = m2g
µsm1 + µsm3 = m2
m3 = `f_s^max` – m1
(ii) Given,
m1 = 15 kg, m2 = 10 kg : m3 = 25 kg and µs = 0.2
m3 = `f_s^max` – m1
m3 = `10/0.2` – 15 = 50 – 15 = 35 kg
The minimum mass m3 = 35 kg has to be placed on ml to prevent it from sliding. But here m3 = 25 kg only.
The combined masses (m1 + m3) will slide.
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