Question

a block of mass ( m_1 ) rests on a rough horizontal tabletop, as shown in the figure. the block is connected by a string to a second block of mass ( m_2 ), which hangs below a pulley at the edge of the table. the coefficient of static friction between the tabletop and the first block is ( mu_s ). the masses of the string and the pulley are negligible, and the pulley can rotate with negligible friction in its axle. what is the minimum mass ( m_2 ) that will cause the blocks to start moving?
a ( m_1 )
b ( mu_s m_1 )
c ( \frac{m_1}{mu_s} )
d ( (1 + mu_s)m_1 )

Explanation:

Step1: Analyze forces on \(m_1\)

The normal force \(N\) on \(m_1\) is \(N = m_1g\) (since it's on a horizontal surface and vertical forces balance). The maximum static - friction force \(f_s=\mu_sN=\mu_sm_1g\).

Step2: Analyze forces on \(m_2\)

When the blocks are on the verge of moving, the tension \(T\) in the string is equal to the weight of \(m_2\), i.e., \(T = m_2g\). Also, at the verge of motion, the tension \(T\) is equal to the maximum static - friction force \(f_s\) (because the tension is trying to pull \(m_1\) and static friction is opposing the motion). So, \(m_2g=\mu_sm_1g\).

Step3: Solve for \(m_2\)

Divide both sides of the equation \(m_2g=\mu_sm_1g\) by \(g\) (since \(g
eq0\)). We get \(m_2=\mu_sm_1\).

Answer:

B. \(\mu_sm_1\)