Comment:
Migration of unmigrated content due to installation of a new plugin
Excerpt
hidden
true
Classic example of static friction on a moving surface.
Composition Setup
Deck of Cards
id
bigdeck
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
label
Part 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
id
sysa
System:
Cloak
id
sysa
We will treat the box as a .
Toggle Cloak
id
inta
Interactions:
Cloak
id
inta
External influence from the earth (gravity) and the truck bed (normal force and friction force).
Toggle Cloak
id
moda
Model:
Cloak
id
moda
.
Toggle Cloak
id
appa
Approach:
Cloak
id
appa
Toggle Cloak
id
diaga
Diagrammatic Representation
Cloak
id
diaga
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!
Cloak
diaga
diaga
Toggle Cloak
id
matha
Mathematical Representation
Cloak
id
matha
Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:
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?
System: As defined in Part A. We will also have to model the truck as a separate [point particle] in order to determine the time for the box to fall off.
Models: [Point Particle Dynamics] and [One-Dimensional Motion with Constant Acceleration].
Approach: We begin with the same analysis as in Part A. Once again, friction is trying to keep the box moving at the same rate as the truck bed. This time, however, in order to accelerate the box at 4.0 m/s ^2^, the friction force would have to have a size of 60 N. We found in Part A that the maximum possible static friction force on the box is 59 N. Thus, the box cannot accelerate at the same rate as the truck and will begin to slide.
Once we have determined the box will begin to slide along the truck bed, we must recognize we are in the kinetic friction regime. The box's acceleration, then, will be:
{latex}\begin{large} \[ a_{x} = \frac{F_{f}}{m} = \frac{\mu_{k}N}{m} = \mu_{k}g = 2.45\:{\rm m/s}^{2} \]\end{large}{latex}
This acceleration is less than the truck's, and so the box will fall behind. To determine when the box falls off the truck, we must now use the equations of one-dimensional kinematics. Both the truck and the box have their own set of state variables and their own Law of Change. Since we are interested in time, we use:
{latex}\begin{large}\[x_{\rm truck} = x_{\rm truck,i} + v_{\rm truck} t + \frac{1}{2} a_{\rm truck} t^{2} \]
\[x_{\rm box} = x_{\rm box,i} + v_{\rm box} t + \frac{1}{2} a_{\rm box} t^{2} \]\end{large}{latex}
We can make these equations simplest by choosing the initial positions to be zero, and we also know that both the box and the truck began at rest. Thus, the equations become:
{latex}\begin{large}\[x_{\rm truck} = \frac{1}{2} a_{\rm truck} t^{2} \]
\[x_{\rm box} = \frac{1}{2} a_{\rm box} t^{2} \]\end{large}{latex}
limit.
Cloak
matha
matha
Cloak
appa
appa
Card
label
Part 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
id
sysintb
System and Interactions:
Cloak
id
sysintb
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
id
modb
Models:
Cloak
id
modb
and One-Dimensional Motion with Constant Acceleration.
