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Part

A

A

15

kg

box

is

sitting

in

the

bed

of

a

pickup

truck.

The

truck

begins

to

accelerate

at

a

constant

rate

of

3.5

m/s². Given that the friction between the box and the truck bed is characterized by a coefficient of kinetic friction of 0.25 and a coefficient of static friction equal to 0.40, what is the magnitude of the friction force acting on the box once the truck begins its acceleration?

Solution

We begin with a sketch that represents the situation, and then create the appropriate free body diagram.

Image Added

Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:

\begin{large}\[\sum F_{x} = F_{f} = ma_{x} \]\[\sum F_{y} = N - mg = ma_{y} \] \end{large}

Since

the

truck

is

moving

only

in

the

x-direction,

we

expect

a

_y =

0.

Thus,

we

know

that

the

normal

force

acting

on

the

box

will

equal

its weight.

If

friction

is

adequate,

we

expect

that

the

box

will

accelerate

in

the

x-direction

at

the

same

rate

as

the

truck

does.

In

that

case,

we

expect:

\begin{large}\[ F_{f} = (15\:{\rm kg})(3.5\:{\rm m/s}^{2}) = 53\:{\rm N} \]\end{large}

It

is

important

now

to

check

that

this

result

does

not

conflict

with

the

requirement

that

\begin{large} \[ F_{f} \le \mu_{s} N\] \end{large}

Since

we

have

already

used

the

y-direction

equation

of

Newton's

Second

Law

to

conclude

that

the

normal

force

on

the

box

is

equal

to

mg

,

we

find:

\begin{large} \[ F_{f} = 53\:{\rm N} < \mu_{s} N = 0.40(15\:{\rm kg})(9.8\:{\rm m/s}^{2}) = 59\:{\rm N} \]\end{large}

We

therefore

conclude

that

our

answer,

F

_f =

53

N,

is

compatible

with

the

static

friction

limit.