Notice, for condition of static friction between the blocks of masses [tex]m_2[/tex] and [tex]m_3[/tex] , the block of mass [tex]m_2[/tex] has to be stationary w.r.t. block of mass [tex]m_3[/tex] which implies [tex]m_2[/tex] moves with horizontal acceleration [tex]a[/tex].
Now, consider free body diagram of both the blocks [tex]m_1 \ \text{and}\ m_2[/tex] as shown in the figure below.
Answers & Comments
Answer:
μ=m₂a+m₁g/m₂g
Explanation:
First u need to calculate the force towards the moving direction of the mass.
M₃ is moving with the constant acceleration of a, that acceleration also applied to the M₂, So
F=M₂a
and the tension created by M₁ mass is also towards the direction of the mass moving
T = M₁g
add the all forces that have same direction
F= M₂a+M₁g
Frictional force should be equal to the total force that we calculated.
Fr=M₂a+M₁g
Fr=μM₂g
μ=M₂a+M₁g/M₂g
Answer:
Explanation:
Notice, for condition of static friction between the blocks of masses [tex]m_2[/tex] and [tex]m_3[/tex] , the block of mass [tex]m_2[/tex] has to be stationary w.r.t. block of mass [tex]m_3[/tex] which implies [tex]m_2[/tex] moves with horizontal acceleration [tex]a[/tex].
Now, consider free body diagram of both the blocks [tex]m_1 \ \text{and}\ m_2[/tex] as shown in the figure below.
[tex]T-m_1g=0\tag 1\\\implies T=m_1g[/tex]
[tex]T-\mu m_2g=m_2a\\\\\implies m_1g-\mu m_2g=m_2a\\\\\mu m_2g=m_1g-m_2a\\\\\mu =\dfrac{m_1g-m_2a}{m_2g}[/tex]