Question

A shipping container breaks away from a flat bed trailer while being transported at a speed of 80 km/hr on an uphill segment of I-24 having a 5.7% grade. Determine the stopping distance in meters if the COF between the road and container is 0.2.

Please show steps!

Answer 1

Initial velocity of shipping container = V₀ = 80 km/hr

V₀ = 80 x 1000/3600 m/s

V₀ = 22.22 m/s

Final velocity of shipping container = V₁ = 0 m/s (the shipping container comes to a stop)

Coefficient of friction = = 0.2

Grade of slope = 5.7%

Tan = 5.7/100

= 3.26^o

Mass of shipping container = m

Deceleration of shipping container = a

Stopping distance of shipping container = d

From the free body diagram

N = mgCos

f = N

f = mgCos

ma = mgSin + f

ma = mgSin + mgCos

a = gSin + gCos

a = 9.81Sin(3.26) + 0.2x9.81xCos(3.26)

a = 2.52 m/s²

As the shipping container is decelerating that is the acceleration and velocity are in opposite directions, we will be taking velocity as positive and acceleration as negative.

a = -2.52 m/s²

V₁² = V₀² + 2ad

0² = 22.22² + 2(-2.52)d

5.04d = 493.73

d = 97.96 m

The stopping distance of the shipping container is 97.96 meters.