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A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level 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 m/s without causing the box to slide? Draw a free-body diagram of the toolbox.

Short Answer

Expert verified
The shortest time is approximately 4.70 seconds.

Step by step solution

01

Identify the Forces Acting on the Toolbox

To solve this problem, first consider the forces acting on the toolbox. The forces include the force of gravity pulling it downwards, the normal force from the truck bed acting upwards, and the static frictional force acting horizontally to prevent it from sliding.- Gravitational Force = \(mg\)- Normal Force = \(N\)- Maximum Static Frictional Force = \(f_s = \mu_s N\)
02

Establish the Conditions for Static Friction

The toolbox will not slide if the static friction force can resist the horizontal force due to acceleration. Since the maximum static frictional force is \(f_s = \mu_s mg\), this force must be equal or greater than the force required to move the toolbox at the truck's acceleration:\(f_s \geq ma\)So, \(\mu_s mg \geq ma\)
03

Calculate Maximum Possible Acceleration

To find the maximum acceleration \(a\) without the toolbox sliding, rearrange the inequality from Step 2:\[\mu_s mg = ma\]Simplifying gives:\[a = \mu_s g\]Plugging in \(\mu_s = 0.650\) and \(g = 9.81 \, \text{m/s}^2\), we can calculate:\[a = 0.650 \times 9.81 = 6.3765 \, \text{m/s}^2\]
04

Calculate the Shortest Time to Reach Desired Speed

Using the formula for constant acceleration, we know:\[v = at\]where \(v = 30.0 \, \text{m/s}\), and solving for \(t\) gives:\[t = \frac{v}{a} = \frac{30.0}{6.3765} \approx 4.70 \, \text{seconds}\]
05

Construct the Free-Body Diagram

Draw a box representing the toolbox. Indicate the force of gravity acting downwards, the normal force acting upwards, the maximum static frictional force acting horizontally to the left, resisting the movement to the right.

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Key Concepts

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

Static Friction
Static friction acts between two surfaces that are not moving relative to each other. It is the force that keeps the toolbox from sliding off the truck when the truck accelerates. Static friction must be overcome for an object to start moving.
The force of static friction depends on:
  • The nature of the surfaces (roughness, etc.)
  • The normal force (weight of the object)
It can be calculated using the formula:
  • Maximum static frictional force: \( f_s = \mu_s N \)
  • \( \mu_s \): coefficient of static friction
  • \( N \): normal (support) force
The concept is crucial because it determines whether the toolbox will slide off or remain on the truck as it accelerates. If the force from acceleration exceeds static friction, the toolbox will slide.
Free-Body Diagram
A free-body diagram is a simple graphical illustration used to visualize the forces acting on an object.
In the context of the toolbox:
  • Draw a box to represent the toolbox
  • Show the gravitational force (\( mg \)) acting downwards
  • Show the normal force (\( N \)) acting upwards
  • Indicate the static frictional force horizontally opposing potential sliding
This diagram helps in analyzing the problem by visualizing all forces involved. It simplifies the complex interactions into a clear picture, making it easier to apply Newton's laws of motion.
By balancing the forces, we can determine if and when the box will start moving due to the truck's acceleration.
Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate. This means the speed increases (or decreases) steadily over time.
The key characteristics include:
  • Constant acceleration value
  • Linear velocity change
In the exercise, the truck accelerates uniformly from rest to 30.0 m/s.
Use the formula:
\[v = at\]
where
  • \( v \): final velocity
  • \( a \): acceleration
  • \( t \): time
Here, you can calculate the shortest time the truck requires to reach a certain speed without the toolbox sliding, provided the static friction force is not exceeded. By understanding the principle of uniform acceleration, you can apply it to find how external forces, like static friction, affect the movement of objects on moving surfaces.

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Most popular questions from this chapter

A 70-kg person rides in a 30-kg cart moving at 12 m/s at the top of a hill that is in the shape of an arc of a circle with a radius of 40 m. (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart can travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

A 2.00-kg box is suspended from the end of a light vertical rope. A time- dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to \(v(t) =\) (2.00 m/s\(^2\))\(t\) \(+\) (0.600 m/s\(^3\))\(t^2\). What is the tension in the rope when the velocity of the box is 9.00 m/s?

Two objects, with masses 5.00 kg and 2.00 kg, hang 0.600 m above the floor from the ends of a cord that is 6.00 m long and passes over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00-kg object.

A box with mass 10.0 kg moves on a ramp that is inclined at an angle of 55.0\(^\circ\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_k =\) 0.300. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F =\) 120.0 N that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

Two crates connected by a rope lie on a horizontal surface (\(\textbf{Fig. E5.37}\)). Crate \(A\) has mass \(m_A\), and crate \(B\) has mass \(m_B\). The coefficient of kinetic friction between each crate and the surface is \(\mu_k\). The crates are pulled to the right at constant velocity by a horizontal force \(F\). Draw one or more free-body diagrams to calculate the following in terms of \(m_A\), \(m_B\), and \(\mu_k\): (a) the magnitude of \(F\) and (b) the tension in the rope connecting the blocks.

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