A baggage conveyor is used to unload luggage from an airplane. The 8-kg duffel bag A is sitting on top of the 18-kg suitcase B. The conveyor is moving the bags down at a constant speed of 0.75 m/s when the belt suddenly stops. Knowing that the coefficient of friction between the belt and B is 0.25 and that bag A does not slip on suitcase B, determine the smallest allowable coefficient of static friction between the bags.

μ = 0.8125

Explanation:

For this problem we use the transactional equilibrium equation

on the vertical axis

        N -W₁ -W₂ = 0

        N = W₁ + W₂

        N = (m₁ + m₂) g

where m₁ and m₂ are the masses of the bag and the suitcase, respectively

the friction force has the expression

        fr = μ N

        fr = μ (m₁ + m₂) g

        fr = 0.25 (8 + 18) 9.8

         fr = 63.7 N

This is the friction force between the suitcase and the conveyor belt. This is the force that keeps the bag moving, let's call it the

         F = fr = 63.7 N

Now let's analyze the situation between the bag and the suitcase, the only force there is is the friction, which in this case goes backwards since the systems stopped

we apply Newton's second law

          F -fr = 0

          F = fr

force of the bag is

          fr = μ N_bag

           fr = μ m_bag  g

 we substitute

          F = μ m_bag  g

          μ = F / (m_bag g)

          μ = 63.7 / (8 9.8)

          μ = 0.8125

the answer is b ) square root of 5