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
Attached Figure / Diagram
A
m1/m2
B
m1g/m2
C
m2/m1
D
m2g/m1
Verified Correct Answer: C.
Physical Principles & Numerical Solution
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 - \mu \(m_1\) g = 0, so that T = \mu \(m_1\)g ---------- 2
equating 1 and 2 above;
Therefore \(m_2\)g = \mu\(m_1\)g
\mu = \(\frac{m_2}{m_1}\)
Source: Myschool Past Questions Archive (Record #41705)