Question

b1. in the system shown above, the block of mass ( m_1 ) is on a rough horizontal table. the string that attaches it to the block of mass ( m_2 ) passes over a frictionless pulley of negligible mass. the coefficient of kinetic friction ( mu_k ) between ( m_1 ) and the table is less than the coefficient of static friction ( mu_s ).
a on the diagram below, draw and identify all the forces acting on the block of mass ( m_1 ).
b. in terms of ( m_1 ) and ( m_2 ) determine the minimum value of ( mu_s ) that will prevent the blocks from moving.
the blocks are set in motion by giving ( m_2 ) a momentary downward push. in terms of ( m_1 ), ( m_2 ), ( mu_k ), and ( g ), determine each of the following:
c. the magnitude of the acceleration of ( m_1 )
d the tension in the string.

Explanation:

Step1: Analyze forces on \(M_1\)

For block \(M_1\), the forces are:

• Gravitational force \(F_{g1}=M_1g\) (downward)
• Normal force \(F_N\) (upward, perpendicular to the table)
• Tension force \(T\) (to the right, from the string)
• Kinetic friction force \(F_f=\mu_kF_N\) (to the left, opposing motion)

Step2: Find minimum \(\mu_s\) (part b)

When the blocks are on the verge of moving (static equilibrium), the tension \(T = M_2g\) (from \(M_2\)'s equilibrium). Also, the maximum static - friction force \(F_{f,s}=\mu_sF_N\) and \(F_N = M_1g\) (vertical equilibrium of \(M_1\)). Since \(F_{f,s}=T\), we have \(\mu_sM_1g = M_2g\).

Step3: Calculate acceleration (part c)

For \(M_2\): \(M_2g - T = M_2a\)
For \(M_1\): \(T-\mu_kM_1g = M_1a\)
Add the two equations: \(M_2g-\mu_kM_1g=(M_1 + M_2)a\)
So, \(a=\frac{M_2g-\mu_kM_1g}{M_1 + M_2}\)

Step4: Calculate tension (part d)

From \(M_1\)'s equation \(T=\mu_kM_1g+M_1a\)
Substitute \(a = \frac{M_2g-\mu_kM_1g}{M_1 + M_2}\)

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Answer:

• Part b: \(\mu_s=\frac{M_2}{M_1}\)
• Part c: \(a=\frac{M_2g-\mu_kM_1g}{M_1 + M_2}\)
• Part d: \(T=\frac{M_1M_2g(1+\mu_k)}{M_1 + M_2}\)