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Chapter 5: Problem 79

A \(30.0-\mathrm{kg}\) packing case is initially at rest on the floor of a \(1500-\mathrm{kg}\) pickup truck. The coefficient of static friction between the case and the truck floor is \(0.30,\) and the coefficient of kinetic friction is \(0.20 .\) Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 \(\mathrm{m} / \mathrm{s}^{2}\) northward and (b) when it accelerates at 3.40 \(\mathrm{m} / \mathrm{s}^{2}\) southward.

Short Answer

Expert verified
(a) 66.0 N northward, (b) 58.8 N southward.

Step by step solution

01

Calculate Normal Force

The normal force \( F_N \) is the force exerted by the floor of the truck on the case. Since the case is on a flat surface, and the only vertical force is due to gravity, the normal force \( F_N \) is equal to the weight of the case.\[ F_N = m_{ ext{case}} \cdot g = 30.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 294 \, \text{N} \]
02

Identify Maximum Static Friction

The maximum static friction \( f_s^{ ext{max}} \) is given by the product of the coefficient of static friction \( \mu_s \) and the normal force \( F_N \).\[ f_s^{ ext{max}} = \mu_s \cdot F_N = 0.30 \cdot 294 \, \text{N} = 88.2 \, \text{N} \]
03

Calculate Required Friction Force for Northward Acceleration

For part (a), the truck accelerates northward at \( a = 2.20 \, \text{m/s}^2 \). The friction force \( f \) required to prevent the case from sliding is calculated using Newton's second law:\[ f = m_{ ext{case}} \cdot a = 30.0 \, \text{kg} \cdot 2.20 \, \text{m/s}^2 = 66.0 \, \text{N} \]
04

Compare with Static Friction and Determine Type of Friction (Northward)

Since \( 66.0 \, \text{N} < 88.2 \, \text{N} \), static friction is sufficient to keep the case from sliding. Thus, the friction force is \( 66.0 \, \text{N} \) northward.
05

Calculate Required Friction Force for Southward Acceleration

For part (b), the truck accelerates southward at \( a = 3.40 \, \text{m/s}^2 \). The required friction is:\[ f = m_{ ext{case}} \cdot a = 30.0 \, \text{kg} \cdot 3.40 \, \text{m/s}^2 = 102.0 \, \text{N} \]
06

Compare with Static Friction and Use Kinetic Friction if Necessary (Southward)

Since \( 102.0 \, \text{N} > 88.2 \, \text{N} \), static friction is not sufficient, so the case will slide. The kinetic friction force \( f_k \) is used:\[ f_k = \mu_k \cdot F_N = 0.20 \cdot 294 \, \text{N} = 58.8 \, \text{N} \]The kinetic friction force is therefore \( 58.8 \, \text{N} \), southward.

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Key Concepts

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

Static Friction
Static friction is the force that keeps an object stationary when a force is applied. It acts in the opposite direction to the applied force up to its maximum limit.
For our exercise, the static frictional force prevents the packing case from sliding as the truck accelerates.
  • The coefficient of static friction, denoted as \( \mu_s \), is the proportionality constant that indicates how strongly two surfaces grip each other.
  • In the problem, \( \mu_s = 0.30 \). This value tells us that the force needed to overcome static friction must be at least 30% of the object's normal force.
  • To find the maximum possible static friction \( f_s^{\text{max}} \), multiply \( \mu_s \) by the normal force \( F_N \).
When the truck accelerates northward at \( 2.20 \, \text{m/s}^2 \), the static friction manages to keep the case stationary and provides a force of \( 66.0 \, \text{N} \), which is less than the maximum static friction available. Thus, the case doesn鈥檛 move.
Kinetic Friction
Kinetic friction occurs when two surfaces slide past one another. It is usually less than static friction, meaning once an object starts moving, less force is required to keep it moving.
In part (b) of our exercise, the pickup truck accelerates southward at \( 3.40 \, \text{m/s}^2 \).
  • Here, static friction is insufficient鈥擻( 102.0 \, \text{N} \) is needed but only \( 88.2 \, \text{N} \) is available. Consequently, the case can no longer remain stationary.
  • The coefficient of kinetic friction, \( \mu_k \), determines the frictional force available once movement begins.
  • For this case, \( \mu_k = 0.20 \). Multiply this by the normal force \( F_N \) to get the kinetic frictional force \( f_k = 58.8 \, \text{N} \).
Therefore, once the case starts to slide, the kinetic friction acts in the southward direction, opposing the acceleration.
Newton's Second Law
Newton鈥檚 second law of motion shows how force, mass, and acceleration relate: \( F = m \cdot a \).
This means that the force acting on an object equals its mass multiplied by its acceleration.
  • In our scenario, we used this law to determine the frictional force necessary to counteract the truck's acceleration鈥攂oth northward and southward.
  • For example, northward acceleration of the truck required a friction force for the \( 30.0 \, \text{kg} \) case, calculated as \( 66.0 \, \text{N} \) using \( F = m_{\text{case}} \cdot a \).
  • Likewise, for southward acceleration, you calculate the needed frictional force as \( 102.0 \, \text{N} \).
By comparing these values with what static and kinetic friction can provide, we determine the type of friction acting to prevent or allow sliding.
Normal Force
Normal force is the support force exerted by a surface, perpendicular to the surface. It is crucial in determining frictional forces because frictional force is usually proportional to the normal force.
In our exercise, since there are no vertical accelerations, the normal force equals the gravitational force on the case.
  • This is calculated as \( F_N = m_{\text{case}} \cdot g = 294 \, \text{N} \), where \( g = 9.8 \, \text{m/s}^2 \) for Earth's gravity.
  • The normal force allows us to compute both static and kinetic friction鈥攚hich are integral when analyzing motion and rest scenarios.
Always remember that the normal force can change if there are additional vertical forces or inclines involved, which would affect friction too.

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Most popular questions from this chapter

A model airplane with mass 2.20 \(\mathrm{kg}\) moves in the \(x y\) -plane such that its \(x-\) and \(y\) -coordinates vary in time according to \(x(t)=\alpha-\beta t^{3}\) and \(y(t)=y t-\delta t^{2},\) where \(\alpha=1.50 \mathrm{m}, \beta=\) \(0.120 \mathrm{m} / \mathrm{s}^{3}, \gamma=3.00 \mathrm{m} / \mathrm{s},\) and \(\delta=1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate the \(x-\) and \(y\) -components of the net force on the plane as functions of time. (b) Sketch the trajectory of the airplane between \(t=0\) and \(t=3.00 \mathrm{s}\) , and draw on your sketch vectors showing the net force on the airplane at \(t=0, t=1.00 \mathrm{s}, t=2.00 \mathrm{s},\) and \(t=3.00 \mathrm{s}\) . For each of these times, relate the direction of the net force to the direction that the airplane is turning, and to whether the airplane is speeding up or slowing down (or neither). (c) What are the magnitude and direction of the net force at \(t=3.00 \mathrm{s} ?\)

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