In the diagram above, the hanging mass \(m_2\) is adjusted until \(m_1\) is on the verge of sliding. The coefficient of static friction between the mass \(m_1\) and the table is
- m1/m2
- m1g/m2
-
m2/m1
- m2g/m1
The correct answer is: C
Explanation
Since the system is on the verge of moving, let the resultant force on \(m_1\) and \(m_2\) be F = 0
so, \(m_2\): F = \(m_2\)g - T = 0 so that \(m_2\)g = T ---------- 1
for \(m_1\) : F = T - μ \(m_1\) g = 0, so that T = μ \(m_1\)g ---------- 2
equating 1 and 2 above;
Therefore \(m_2\)g = μ\(m_1\)g
μ = \(\frac{m_2}{m_1}\)