In figure, coefficient of friction between m1 and m2 is m and that between m1 and the wall is zero. A force F is pressing the system against the wall. Minimum value of force required to hold the system in equilibrium is A.mm1g B.mm2gC.m(m1 + m2)g D.system cannot be held in equilibrium
Question
In figure, coefficient of friction between m1 and m2 is m and that between m1 and the wall is zero. A force F is pressing the system against the wall. Minimum value of force required to hold the system in equilibrium is A.mm1g B.mm2gC.m(m1 + m2)g D.system cannot be held in equilibrium
Solution
The system can be held in equilibrium if the frictional force between m1 and m2 is equal to the weight of m2. The frictional force is given by the equation F_friction = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is the force F pressing the system against the wall.
The weight of m2 is given by the equation W = m2g, where g is the acceleration due to gravity.
Setting these two equations equal to each other gives us:
μN = m2g
Substituting F for N gives us:
μF = m2g
Solving for F gives us:
F = m2g/μ
Since the coefficient of friction between m1 and m2 is m, we substitute m for μ to get:
F = m2g/m
Therefore, the minimum value of force required to hold the system in equilibrium is m2g/m, which corresponds to option B. mm2g.
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