(a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is notstrapped down, in violation with USDOT regulations. If the coefficient of static friction between steel andwood is 0.52, what is the minimum distance over which the truck can come to a stop without the girdersliding toward the cab of the truck? (answer: 77 m)(b) What is the minimum time over which the truck can accelerate forward from 0 m/s to 28 m/s with aconstant acceleration without the girder sliding off the back? (answer: 5.5 s)FNET = mafs.max = μsnW = mgv² = v₁² + 2aAxV = Vo+ atg=9.81 m/s²

Name: (a) A flatbed truck moving at 28 m/s carries a steel girder that rest
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Description: (a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is notstrapped down, in violation with USDOT regulations. If the coefficient of static friction between steel andwood is 0.52, what is the minimum

(a) Given that, Initial speed of the truck () = 28 m/s…

Answered: (a) A flatbed truck moving at 28 m/s…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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(a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is not
strapped down, in violation with USDOT regulations. If the coefficient of static friction between steel and
wood is 0.52, what is the minimum distance over which the truck can come to a stop without the girder
sliding toward the cab of the truck? (answer: 77 m)
(b) What is the minimum time over which the truck can accelerate forward from 0 m/s to 28 m/s with a
constant acceleration without the girder sliding off the back? (answer: 5.5 s)
FNET = ma
fs.max = μsn
W = mg
v² = v₁² + 2aAx
V = Vo+ at
g=9.81 m/s²

Expert Solution

Step 1: Determine the given variables

(a) Given that, Initial speed of the truck ( $v subscript 0$ ) = 28 m/s

Coefficient of static friction ( $mu subscript s$ ) = 0.52

Final speed ( $v$ ) = 0m/s (when the truck stops)

Net force on the truck , $F subscript N E T end subscript space equals space m a$ --------Eqn(1)

Maximum frictional force , $space f subscript s comma m a x end subscript space equals space mu subscript s N$ -------Eqn(2)

where, N is the normal force and is given by,

$N space equals space m g$

substituting in Eqn(2) gives,

$f subscript s comma m a x end subscript equals space mu subscript s m g$ -------Eqn(3)

when the truck is going to stop, Net force on the truck is equal to the maximum frictional force. Then, Equating Eqn(1) and Eqn(3),

$space m a space equals space mu subscript s m g$

Gives, Acceleration $a space equals space mu subscript s g$ ------Eqn(4)

where, g - Acceleration due to gravity ( g = 9.81 m/s²)

Substituting the values in Eqn(4)

$a space equals space left parenthesis 0.52 right parenthesis cross times left parenthesis 9.81 m divided by s squared right parenthesis a space equals space 5.1012 space m divided by s squared$

When the truck is going to stop, then the acceleration is retarding

i.e, $bold italic a bold space bold equals bold space bold minus bold 5 bold. bold 1012 bold space bold italic m bold divided by bold italic s to the power of bold 2$

Now, using the kinematic equation

$v squared space equals space v subscript 0 superscript 2 space plus space 2 a increment x$

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