Question

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed of the rigis 72 km/h, determine the shortest distance in which the rig can bebrought to a stop if the load is not to shift

Answer 1

Answer:

50.97 m

Explanation:

m = Mass of truck

$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.