Question

A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at constant speed

Answer 1

• The horizontal pushing force required to just start the crate moving is 447 N.

• The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

Explanation:

• By the definition of the coefficient of static friction we have:

                                 μ$_{s}$ = $\frac{F_{appl} }{W}= \frac{F_{s} }{N}$,

where, $F_{appl}$ is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            $F_{s}$ is  the static  friction  force-directed  opposite  to  the  horizontal  pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From  this  formula  we  can  find the horizontal pushing  force  required to  just  start the crate moving:

                         $F_{appl} = F_{s} = u_{s}N = u_{s}mg$

                                                      = 0.760 $\times$ 60 kg $\times$ 9.8 m / s^2

                                                      = 447 N.

• By the definition of the coefficient of kinetic friction we have:

                              u$_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}$,

where, $F_{appl}$ is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            $F_{k}$ is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

                              $F_{appl} = F_{k} = u_{k}N = u_{k}mg$

                                                            = 0.410 $\times$ 60 $\times$ 9.8

                                                            = 241 N.

• The horizontal pushing force required to just start the crate moving is 447 N.

• The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.