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Excerpt
hiddentrue

Classic example of static friction on a moving surface.

Composition Setup
 Deck of Cards
idbigdeck
 Wiki Markup
h3. 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/s ^2^.  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?

System:  We will treat the box as a [point particle] subject to external influence from the earth (gravity) and the truck bed (normal force _and_ friction force).  

Model:  [Point Particle Dynamics].

Approach:  We begin with a sketch that represents the situation, and then create the appropriate free body diagram.
 
!closethegate.png!

{note}It is important to note that friction works to prevent movement of the interface between the box and the truck bed.  The truck bed is moving forward, so friction will attempt to pull the box forward as well.  If the box moves at the same rate as bed, then the interface is static.  For this reason, "static" friction will actually _cause_ motion of the box in this case!{note}

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

{latex}
 Card
labelPart A
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
 Toggle Cloak
idsysa
 System: 
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idsysa
We will treat the box as a .
 Toggle Cloak
idinta
 Interactions: 
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idinta
External influence from the earth (gravity) and the truck bed (normal force and friction force).
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
.
 Toggle Cloak
idappa
 Approach:
 Cloak
idappa
 Toggle Cloak
iddiaga
 Diagrammatic Representation
 Cloak
iddiaga
We begin with a sketch that represents the situation, and then create the appropriate free body diagram.
Image Added
 Note
It is important to note that friction works to prevent movement along the interface between the box and the truck bed. The truck bed is moving forward, so friction will attempt to pull the box forward as well. If the box moves at the same rate as bed, then the interface is static. For this reason, "static" friction will actually cause motion of the box in this case!
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diaga
diaga
 Toggle Cloak
idmatha
 Mathematical Representation
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idmatha
Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:
 Latex
\begin{large}\[\sum F_{x} = F_{f} = ma_{x} \]\[\sum F_{y} = N - mg = ma_{y} \] \end{large}
{latex}


Since 
 
 the 
 
 truck 
 
 is 
 
 moving 
 
 only 
 
 in 
 
 the 
 
 x-direction, 
 
 we 
 
 expect 
 _
a
_~y~ 
y = 
 
 0. 
 
 Thus, 
 
 we 
 
 know 
 
 that 
 
 the 
 
 normal 
 
 force 
 
 acting 
 
 on 
 
 the 
 
 box 
 
 will 
 
 equal 
 it 
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:


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

We are not finished yet.  It is important now to check that this result does not conflict with the requirement that

{latex}
 Warning
We're not finished yet!
It is important now to check that this result does not conflict with the requirement that
 Latex
\begin{large} \[ F_{f} \le \mu_{s} N\] \end{large}
{latex}


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:


{
 Latex
}
\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}
{latex}


We 
 
 therefore 
 
 conclude 
 
 that 
 
 our 
 
 answer, 
 _
F
_~f~ 
f = 
 
 53 
 
 N, 
 _
is
_ 
 compatible 
 
 with 
 
 the 
 
 static 
 
 friction 
 limit.

h3. Part B

Consider the same basic situation as above, but now suppose that the truck accelerates at a rate of 4.0 m/s ^2^ rather than 3.5 m/s ^2^. If the center of mass of the box is located 2.0 m horizontally from the edge of the truck bed, how much time will elapse from the instant the truck begins to accelerate until the instant the box falls off the truck bed?

Approach:  We use the same system and model as in Part A.  Once again, friction is trying to keep the box moving at the same rate as the truck bed.



limit.
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matha
matha
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appa
appa
 Card
labelPart B
Part B
Consider the same basic situation as above, but now suppose that the truck accelerates at a rate of 4.0 m/s2 rather than 3.5 m/s2. If the center of mass of the box is located 2.0 m horizontally from the edge of the truck bed, how much time will elapse from the instant the truck begins to accelerate until the instant the box falls off the truck bed?
Solution
 Toggle Cloak
idsysintb
 System and Interactions: 
 Cloak
idsysintb
As defined in Part A. We will also have to model the truck as a separate  in order to determine the time for the box to fall off.
 Toggle Cloak
idmodb
 Models: 
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idmodb
 and One-Dimensional Motion with Constant Acceleration.
 Toggle Cloak
idappb
 Approach: 
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idappb