Two masses `m_(1)` and `m_(2)` are connected with a string passing over a frictionless pulley fixed at the corner of the table as shown in the the coefficent of static friction of mass `m_(1)` with the table is `mu_(s)` calculate the minium mass `m_(3)` that may be placed on `m_(1)` to prevent it from sliding check if `m_(1)=15 kg m_(2)=10kg , mm_(3)=25` and `mu_(s)=0.2`

Explanation

Let` m_3` is the mass added on `m_1`
Maximal static friction
`f_(s)^(max) = mu_s N = mu_s (m_1 + m_3) g ` .....(1)
`N = (m_1 + m_3) g`
`therefore ` Tension acting on string ` = T = m_2g` ....(2)
Equate (1) and (2)
`mu_s (m_1 + m_3) g = m_2 g`
` mu_s m_1 + mu_s m_3 = m_2`
`m_3 = (m_2)/(mu_s) -m_1`
(ii) Given `m_1 = 15kg ,m_2 = 10 kg , m_3 = 25 kg " and " mu_s = 0.2 `
`m_3 = (m_2)/(mu_s) - m_1`
` therefore m_3 = (10)/(0.2) - 15 = 50 -15 = 35 kg `
The minimum mass `m_3` = 35 kg has to be placed on `m_1` to prevent if from sliding . But here `m_3 = 25 kg ` only
` therefore ` The combined mas `(m_1 + m_3) ` will slide .