Question

Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. The train is travelling down a 5 percent grade when it decreases its speed at a constant rate from 144 mi/h to 66 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the baggage car if the trunk is not to slide.

Answer 1

Answer:

 μ = 0.36

Explanation:

This exercise must be solved in parts

Let's start by finding the acceleration of the train, using the kinematics relations

         v = v₀ - a t

         a = $\frac{v - v_o}{t}$

let's reduce the magnitudes to the SI system

         v = 66 mi / h (5280 ft / 1mile) (1h / 3600s) = 96.8 ft / s

         v₀ = 144 mi / h = 211.2 ft / s

let's calculate

          a = $\frac{96.8 - 211.2}{12}$

          a = - 9.53 ft / s²

Now we can use Newton's second law, for this let's set a reference system with the x axis parallel to the slope of the hill

let's break down the weight

            sin 5 = Wₓ / W

            cos 5 = W_y / W

            Wₓ = W sin 5

            W_y = W cos 5

Y axis

     N- W_y = 0

     N = W cos 5

   

X axis

      Wₓ -fr = m a

       mg sin 5 - fr = ma

the friction force has the expression

       fr = μ N

       fr = μ W cos 5

we substitute

        mg sin 5 - μ mg cos 5 = ma

        g sin 5 - μ g cos 5 = a

        μ = $\frac{g \ sin 5 -a }{g \ cos 5}$

        μ =  $tan 5 - \frac{a}{ g \ cos 5}$

let's calculate

       μ = tan 5 - (-9.53) / 32 cos 5

       μ = 0.08749 + 0.2733

       μ = 0.36