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

Here Tension ,
`T= F_(f) = m_(2) g`
`T= mu_(s) (m_(1)+ m_(3))g`
`mu_(s) (m_(1) + m_(3)) g= m_(2)g`
`m_(3) = (m_(2) -mu , m_(1))/( mu)`
`m_(3)= (m_(2))/(mu) = m_(1)`