Question

The coefficent of static friction between the floor of a truck and a box resting on it is 0.28. The truck is traveling at 72.4 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide

Answer 1

75.25 m is the least distance in which the truck can stop and ensure that the box does not slide.

Explanation:

Friction force can withstand the relative movement of hard surfaces, layers of fluids, and material element that slide relative to each other. Dry friction is one type that counteracts the relative lateral movement of two hard surfaces in contact. The equation is given by

                $f=\mu\ N$ --------> eq 1

Where,

f = friction force  

μ = coefficient of friction  

 N = normal force  

Given data:

The coefficient of friction μ  =  0.35

The speed of the truck,  u  =  81.8  k m / h r

after converting it into S.I. unit, u = $\frac{81.8 \times 10^{3}}{3600} \mathrm{m} / \mathrm{s}$ = 22.72 m/s

According to newtons second law of motion, the force applied to the truck is,

                  F = m a -----> eq 2

Where,

m = mass

a = acceleration due to gravity

The force balance equation for limiting condition is,

                              F = f

                          m a = μ N

Substitute N = m g, we get

          $m \times a=\mu \times m \times g$

         $a=\mu \times g$

The third law of motion is,

        $v^{2}=u^{2}+2 a s$

Here, the final velocity is  v .  Hence, the final velocity is zero so  v  =  0 .

By substituting the known values in the above equation, we get

   $0^{2}=u^{2}+2(\mu \times g) s$

   $0^{2}=(22.72)^{2}+(2 \times 0.35 \times(-9.8) \times s)$

   $6.86 \times s = 516.198$

   $s=\frac{516.198}{6.86}=75.25\ \mathrm{m}$