Two blocks m₁ and m₂ are in contact on a horizontal rough floor. Their coefficients of kinetic friction with the floor are μ₁ and μ₂, respectively. A horizontal force F is applied to m₁, pushing both blocks towards m₂. Assuming both slide while remaining in contact, the contact force on m₂ is:
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Question

Two blocks m₁ and m₂ are in contact on a horizontal rough floor. Their coefficients of kinetic friction with the floor are μ₁ and μ₂, respectively. A horizontal force F is applied to m₁, pushing both blocks towards m₂. Assuming both slide while remaining in contact, the contact force on m₂ is:

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G1 Physics · Accepted Answer

For the two-block system, acceleration a = [F - μ₁m₁g - μ₂m₂g]/(m₁+m₂). For block m₂ alone, N - μ₂m₂g = m₂a. Hence N = m₂a + μ₂m₂g. Substituting a and simplifying gives N = [m₂F + m₁m₂g(μ₁ - μ₂)]/(m₁+m₂).