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Chapter 3: Problem 33

The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{f}=\mu \mathrm{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{f}=m a_{x}\). However, if the box is on the verge of slipping, \(F_{f}\) \(=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

Short Answer

Expert verified
The maximum acceleration is 5.9 m/s虏 without the box sliding.

Step by step solution

01

Identify the Problem

We need to find the maximum acceleration of the truck so that the box does not slide along the bed of the truck. Given the coefficient of static friction between the box and the truck bed is \( \mu_s = 0.60 \).
02

Express Friction Force in Terms of Static Friction

When the box is on the verge of slipping, the friction force \( F_f \) equals the maximum static friction force which is \( F_f = \mu_s F_W \). Here, \( F_W \) is the weight of the box.
03

Relate Friction Force to Acceleration

The friction force should also provide the necessary force to accelerate the box. According to Newton's second law, \( F_f = m a_x \), where \( a_x \) is the acceleration of the truck and the box.
04

Equate and Simplify Equations

Since at the verge of slipping \( F_f = \mu_s F_W \), we set \( \mu_s F_W = m a_x \). Given that \( F_W = mg \), the equation becomes: \( \mu_s mg = ma_x \).
05

Solve for Maximum Acceleration

Cancel the mass \( m \) from both sides to get \( a_x = \mu_s g \). Substitute the known values to find the maximum acceleration: \( a_x = 0.60 \times 9.81 \, \text{m/s}^2 \).
06

Calculate and Conclude

Calculate \( 0.60 \times 9.81 = 5.886 \), which we round to \( a_x = 5.9 \, \text{m/s}^2 \). This is the maximum acceleration without the box sliding.

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Key Concepts

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

Newton's second law
Newton's second law of motion is a fundamental principle that describes the relationship between an object's mass, the force applied to it, and its acceleration. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, which is expressed mathematically as:\[ F = ma \]Where:
  • \( F \) is the force acting on the object,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration.
In the context of the exercise, this principle helps us understand that the friction force exerted by the surface must match the force needed to accelerate the box. If this frictional force is greater than or equal to the required force for acceleration, then the box will not slide. When the friction force is the only source of acceleration, using the law simplifies our calculations by linking friction directly to acceleration, helping us solve for unknowns like maximum acceleration.
Maximum acceleration
Maximum acceleration refers to the highest rate of change of velocity an object can have without causing motion changes, such as sliding or slipping. In the case of the box on the truck, we want the truck to accelerate as much as possible without the box sliding off. This is controlled by the force of static friction, which opposes motion initiation.

To find this maximum acceleration, we use the expression:\[ a_{x}=\mu_{s}g \]Where:
  • \( a_x \) is the maximum acceleration,
  • \( \mu_s \) is the coefficient of static friction, and
  • \( g \) is the acceleration due to gravity (approximately 9.81 m/s虏).
By calculating \( \mu_s g \), we find the highest acceleration the truck can achieve while ensuring the box remains stationary relative to the truck's bed. This ensures safety and stability during transportation.
Friction force
Friction force is the resistance to motion that occurs when two surfaces are in contact. In the problem, it's the force preventing the box from sliding off the truck bed as it accelerates. Friction force is crucial in balancing forces so objects don't move unnecessarily.

The friction force is given by:\[ F_{f} = \mu_{s} F_{W} \]Where:
  • \( F_f \) is the friction force,
  • \( \mu_s \) is the coefficient of static friction, and
  • \( F_W \) is the weight of the box.
Friction plays a dual role鈥攊t's necessary to prevent slipping but must not be overwhelmed by other forces that could cause unwanted motion. Therefore, understanding how friction operates and its limitations is key to ensuring the box does not slide during acceleration.
Coefficient of static friction
The coefficient of static friction \( \mu_s \) quantifies the amount of friction between two surfaces that are not in motion relative to each other. It's a dimensionless number that represents the ratio of the maximum static friction force to the normal force.

In the truck and box scenario:
  • \( \mu_s \) tells us how much frictional force the truck bed can exert on the box before it starts to slide.
  • Higher values of \( \mu_s \) mean more friction and, therefore, a greater resistance to the initial motion of the box on the truck.
Understanding \( \mu_s \) helps designers and engineers to predict and control motion. In practical applications, knowing this value allows for the safe design of transportation systems where loads are secure, thus preventing accidents caused by unexpected slippage.

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Most popular questions from this chapter

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

Two forces, \(80 \mathrm{~N}\) and \(100 \mathrm{~N}\), acting at an angle of \(60^{\circ}\) with each other, pull on an object. (a) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

A \(20.0 \mathrm{~kg}\) object that can move freely is subjected to a resultant force of \(45.0 \mathrm{~N}\) in the \(-x\) -direction. Find the acceleration of the object. We make use of the second law in component form, \(\Sigma F_{x}=m a_{x}\), with \(\Sigma F_{x}=-45.0 \mathrm{~N}\) and \(m=20.0 \mathrm{~kg}\). Then $$ a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2} $$ where we have used the fact that \(1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}\). Because the resultant force on the object is in the \(-x\) -direction, its acceleration is also in that direction.

A 600 -kg car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( \(a\) ) First find the car's acceleration from a constant- \(a\) equation. It is known that \(\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0\), and \(x=70 \mathrm{~m}\). Use \(v_{k}^{2}=v_{u}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in \((a)\) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(u=F_{f} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900 -g mass. \((a)\) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. ( \(b\) ) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.
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