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Chapter 5: Problem 23

The Trendelenburg Position. After emergencies with major blood loss, a patient is placed in the Trendelenburg position, in which the foot of the bed is raised to get maximum blood flow to the brain. If the coefficient of static friction between a typical patient and the bed sheets is \(1.20,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

Short Answer

Expert verified
The maximum angle at which the bed can be tilted before the patient begins to slide is approximately \( 50.19^{\circ} \).

Step by step solution

01

Identifying known and unknown values

The coefficient of static friction, denoted generally by \( \mu_s \), is given as \(1.20\). The maximum angle, denoted by \( \Theta \), at which the bed can be tilted is what we're solving for in this case.
02

Set up the static friction equation

We know that the force of static friction \( F_s \) equals the coefficient of static friction times the normal force \( F_n \). And that for an incline scenario, static friction \( F_s \) equals the component of gravity \( mg \) that acts parallel to the incline. Thus we have the following equation.\[ \mu_s F_n = mg \sin(\Theta) \]
03

Replace normal force in the equation

On an incline, the normal force \( F_n \) is equal to the component of the gravitational force \( mg \) that acts perpendicular to the incline, which is \( mg \cos(\Theta) \). Substitute \( F_n \) in the friction equation with this expression, we get:\[ \mu_s mg \cos(\Theta) = mg \sin(\Theta) \]
04

Simplify and isolate the variable

Simplify the equation by canceling out the \( mg \) terms. We are then left with:\[ \mu_s \cos(\Theta) = \sin(\Theta) \]Divide both sides of the equation by \( \cos(\Theta) \), we get:\[ \mu_s = \tan(\Theta) \]Isolate \( \Theta \) by taking the inverse tangent on both sides:\[ \Theta = \arctan(\mu_s) \]
05

Calculate the maximum angle

Substitute the given \( \mu_s \) into the equation to find the value for \( \Theta \).\[ \Theta = \arctan(1.20) = 50.19^{\circ} \]

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Key Concepts

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

Incline Physics
Incline physics plays a vital role in understanding how objects move on inclined planes, such as beds tilted in hospitals. When we tilt a bed, we're essentially creating a slope, and interesting things happen to the forces acting on a patient resting on that slope. The point is to understand how gravity and friction work together.
Gravity pulls the patient downwards, and depending on the angle of incline, part of this gravitational force will try to pull the patient down the slope. This force is what we must counteract to prevent sliding.
Incline physics helps in considering these effects and how the angle of the tilt impacts these forces. Recognizing this is crucial to safe patient positioning and is pertinent in many real-world applications like road designs, construction, and more.
Static Coefficient
The static coefficient of friction is a measure of how resistant a stationary object is to being moved across a surface. In our scenario, the static coefficient is the key value we need to find, as it tells us how much grip the patient has on the bed.

When we talk about the static coefficient ( ) of friction, we describe the ratio of the maximum static frictional force between two surfaces before movement begins, compared to the normal force pressing the surfaces together.
The given static coefficient in this exercise is 1.20, an essential factor in calculating the angle of incline at which sliding begins. This high coefficient implies a considerable amount of friction, meaning the bed can be tilted at a quite steep angle without the patient starting to slide.
  • It's a crucial part of the formula that helps us find out the critical angle for incline-related problems.
  • It varies based on the surface materials in contact. In this case, fabric against fabric or skin.
Gravity Components
The force of gravity acting on any object can be broken into components when we have an inclined plane involved. Understanding these components is crucial to solving problems like the one in this exercise.
Gravity acts downward, but on an incline, it's helpful to break it down into two components:
  • The component that pulls parallel to the plane ( ) 鈥� this tries to move the patient down the slope.
  • The component that acts perpendicular to the plane ( ) 鈥� this is balanced by the normal force, ensuring the patient is pressed onto the bed instead of sinking through it.
By splitting these components, we derive useful equations to work out the angle at which sliding starts. This step is necessary for calculating how different angles affect different forces when dealing with real-world problems involving slopes.

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Most popular questions from this chapter

A box of bananas weighing \(40.0 \mathrm{~N}\) rests on a horizontal surface. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of \(6.0 \mathrm{~N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{~N},\) what is the magnitude of the friction force and what is the box's acceleration?

Traffic Court. You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled \(192 \mathrm{ft}\) before stopping, and the coefficient of kinetic friction between the road and his tires was \(0.750 .\) He was charged with speeding in a \(45 \mathrm{mi} / \mathrm{h}\) zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A racetrack curve has radius \(120.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.300 .\) A race car with mass \(900 \mathrm{~kg}\) rounds the curve with the minimum speed needed to not slide down the banking. (a) As the car rounds the curve, what is the normal force exerted on it by the road? (b) What is the car's speed?

\(\mathrm{A} 50.0 \mathrm{~kg}\) stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(95.0 \mathrm{~m} / \mathrm{s},\) what is the minimum radius of the circle so that the acceleration at this point will not exceed \(4.00 g ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?
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