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Chapter 5: Problem 79

\(\bullet$$\bullet\) A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. The coefficients of kinetic and static friction between the box and the bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 30.0 \(\mathrm{m} / \mathrm{s}(\approx 60 \mathrm{mph})\) without causing the box to slide. Include a free-body diagram of the toolbox as part of your solution. (Hint: First use Newton's second law to find the maximum acceleration that static friction can give the box, and then solve for the time required to reach 30.0 \(\mathrm{m} / \mathrm{s} .\) )

Short Answer

Expert verified
The shortest time to accelerate is approximately 4.71 seconds.

Step by step solution

01

Understand the Problem

We need to find out how quickly a truck can accelerate from rest to 30 m/s without the toolbox sliding due to friction. We'll use the coefficients of static friction to determine the maximum acceleration possible without the toolbox sliding.
02

Use Newton's Second Law

According to Newton's second law, the maximum acceleration \(a_{max}\) can be found using the static friction force \(F_s\), which is given by \( F_s = \mu_s m g \). Here, \( \mu_s = 0.650 \) is the coefficient of static friction, and \( g = 9.8\, \text{m/s}^2 \) is the acceleration due to gravity. Since \( F_s = ma_{max} \), we have \( \mu_s mg = ma_{max} \). Thus, \( a_{max} = \mu_s g \).
03

Calculate Maximum Acceleration

Substitute the values into the equation for maximum acceleration: \[ a_{max} = 0.650 \times 9.8\, \text{m/s}^2 = 6.37\, \text{m/s}^2 \]. This is the maximum acceleration the truck can achieve without the toolbox sliding.
04

Calculate Time to Reach 30 m/s

Using the kinematic equation for constant acceleration: \( v = u + at \), where \( v = 30\, \text{m/s} \), \( u = 0\, \text{m/s} \) (since it starts from rest), and \( a = 6.37\, \text{m/s}^2 \). Rearrange to solve for time \( t \): \[ t = \frac{v - u}{a} = \frac{30\, \text{m/s}}{6.37\, \text{m/s}^2} \approx 4.71\, ext{seconds} \].
05

Free-body Diagram of the Toolbox

Draw a diagram showing the forces acting on the toolbox. Vertically, there's the force of gravity \( F_g = mg \) acting downward and the normal force \( N = mg \) acting upward. Horizontally, static friction \( F_s = \mu_s mg \) acts to the left to counteract the potential sliding of the toolbox caused by the acceleration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction
When you place an object on a surface, and neither are moving, static friction is the force that keeps it in place. It acts to resist any initial motion. In our exercise, static friction is the key player because it holds the toolbox steady on the truck bed while the truck accelerates.
Static friction can be calculated using the formula:\[F_s = \mu_s \times N\]
where:\
    \
  • \(\mu_s\) is the coefficient of static friction.
    • \
  • \(N\) is the normal force, usually equivalent to the object's weight, or \(mg\) in absence of vertical acceleration.
    • \
  • \(m\) is mass.
    • \
  • \(g\) is gravitational acceleration \(9.8\, \text{m/s}^2\).
Static friction is at its maximum just before the object starts moving. It's like the holding force that has to be overcome for motion to start. Once an object begins sliding, static friction gives way to kinetic friction.
Kinetic Friction
Kinetic friction comes into play when two surfaces are sliding past one another. Unlike static friction, kinetic friction acts to slow down the moving object. In our truck example, if the toolbox were to start sliding, kinetic friction would take over from static friction.
The formula for kinetic friction is similar to static friction and is given by:\[F_k = \mu_k \times N\]
where:
  • \(\mu_k\) is the coefficient of kinetic friction.
  • \(N\) is the normal force.
In this exercise, the coefficient of kinetic friction is lower than static friction \((0.355 < 0.650)\). This reflects that it's easier to keep an object moving than to start its motion from rest. However, since we want to avoid the toolbox moving, kinetic friction remains unused here.
Acceleration
Acceleration is how quickly an object changes its velocity. In this context, it's how fast the truck speeds up from rest to its desired speed.
From the exercise, we find the maximum allowable acceleration using Newton's second law and static friction. Using the equation:
\[a_{max} = \mu_s \times g\]
we find the maximum acceleration is \(6.37\, \text{m/s}^2\).
To find how long it takes to reach a speed of \(30\, \text{m/s}\), use the kinematic equation for constant acceleration:
\[v = u + at\]
Rearranging the formula gives us:
\[t = \frac{v-u}{a}\]
Substituting in the known values (\(v = 30 \text{m/s}\) and \(a = 6.37 \text{m/s}^2\)) over time, we calculate that it takes approximately \(4.71\) seconds for the truck to accelerate to the desired speed without the toolbox sliding.

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