Question

A trailer truck with a 2000 [kg] cab and a 8000 [kg] trailer is traveling on a level road at 90 [km/hr].The brakes on the trailer fail, and the anti-skid system of the cab provides the largest possible forcewhich will not cause the wheels of the cab to slide. Knowing that the coefficient of static friction isμs= 0.65, determine (a) the shortest time for the rig to come to a stop, (b) the force in the couplingduring that time.

Answer 1

Answer:

a)   t = 19.6 s, b) fr = 1.274 10⁴ N

Explanation:

This is a Newton's second law problem

Y Axis

for the cabin

        N₁-W₁ = 0

        N₁ = W₁

for the trailer

        N₂- W₂ = 0

        N₂ = W₂

X axis

for the cabin plus trailer, where friction is only in the cabin

         fr = (m₁ + m₂) a

the friction force equation is

        fr = μ N

we substitute

       μ N₁ = (m₁ + m₂) a

        μ m₁ g = (m₁ + m₂) a

        a = μ g    $\frac{m_1}{m_1 + m_2}$

         

let's calculate

         a = 0.65 9.8    $\frac{2000}{2000+8000}$

         a = 1,274 m / s²

a) to find the stopping distance we can use kinematics

         Let's slow down the sI system

         v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s

         v = v₀ - a t

when it is stopped its speed is zero

           0 = v₀ - at

           t = v₀ / a

           t = 25 / 1.274

           t = 19.6 s

b) the friction force is

           fr = 0.65 2000 9.8

          fr = 1.274 10⁴ N

This is the braking force and also the forces that couple the cars.