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JAMB Physics 2018 Paper

Question : 17

Total: 87

In the diagram given the hanging mass m2 is adjusted until m1 is on the verge of sliding. The coefficient of static

Solution:

Concept:

At the verge of sliding, the maximum static friction on

m_1

equals the tension, which is provided by the weight of

m_2

Explanation:

For the hanging mass

m_2

, when it is just supporting the system at rest,

T = m_2g

For

m_1

on the verge of sliding,

f_s = T = m_2g

The normal reaction on

m_1

R = m_1g

The coefficient of static friction is

\mu_s = \frac{f_s}{R} = \frac{m_2g}{m_1g} = \frac{m_2}{m_1}

Answer:

\frac{m_2}{m_1}

i.e. option C.

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