Problem 25 - University Physics with Modern Physics

Question

In emergencies with major blood loss, the doctor will order the patient 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 the 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?

Step-by-Step Solution

Answer

The maximum angle is approximately 50.2 degrees.

Step 1: Understanding the problem

We need to determine the maximum angle at which a bed can be tilted before a patient begins to slide, given that the coefficient of static friction between the patient and the bed sheets is 1.20. This involves using the relationship between static friction force, gravitational force components, and the angle of incline.

Step 2: Identifying the acting forces

The forces acting on the patient are the component of gravitational force pulling the patient down the slope and the static frictional force resisting this motion. The component of gravitational force along the slope is given by \(mg \sin \theta\) and the static frictional force is given by \(f_s = \mu_s N\), where \(N\) is the normal force.

Step 3: Expressing the normal force

The normal force \(N\) is perpendicular to the inclined plane and is given by the component of gravitational force perpendicular to the slope. Thus, \(N = mg \cos \theta\).

Step 4: Setting up the friction condition

For the patient to just start sliding, the maximum static friction force equals the component of the gravitational force along the incline: \(mg \sin \theta = \mu_s mg \cos \theta\).

Step 5: Simplifying the equation

We can cancel \(mg\) from both sides of the equation, giving us \(\sin \theta = \mu_s \cos \theta\). Simplifying further, we get \( \tan \theta = \mu_s\), hence \(\theta = \arctan(\mu_s)\).

Step 6: Calculating the maximum angle

Substitute the given coefficient of static friction, \(\mu_s = 1.20\), to find \(\theta\).\[\theta = \arctan(1.20)\]Use a calculator to find \(\theta\).

Step 7: Finding the angle

Using a calculator, \(\theta = \arctan(1.20) \approx 50.2^\circ\).

Key Concepts

Static Friction

Static friction is the force that keeps objects at rest when they are in contact with a surface, preventing them from sliding. It acts parallel to the contact surface and is a crucial factor when considering objects on inclined planes.
Its magnitude can vary depending on the forces acting upon the object, but it reaches its maximum just before the object begins to move.
This maximum static friction force is given by the formula:

• \[ f_s = \mu_s \, N \]

where \( f_s \) is the static friction force, \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force.
In our problem, the static friction is what prevents the patient from sliding down the inclined bed, and we use this formula to calculate the point where friction no longer holds the patient in place.

Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal. It is a simple machine that reduces the amount of force needed to lift an object by extending the distance over which the force is applied.
In this exercise, the inclined plane is the bed tilted at an angle to help in medical situations.
To analyze problems on an inclined plane, we break down the gravitational force into two components:

• One parallel to the plane, which tries to pull the object down.
• One perpendicular to the plane, which contributes to the normal force.

The angle of the incline significantly affects these components, especially when concerning the balance between gravitational force and static friction.

Gravitational Force

Gravitational force is the force that pulls objects toward each other, keeping everything anchored on Earth's surface.
It acts downward due to the object's mass and Earth's gravity:\[ F_g = mg \]
where \( m \) is the mass and \( g \) is the acceleration due to gravity. On an inclined plane, we consider how this force divides into:

• \( mg \sin \theta \) - the component pulling the patient down the incline.
• \( mg \cos \theta \) - the component contributing to the normal force.

Managing these components helps us understand how gravity affects movement on an inclined plane.

Normal Force

The normal force is the force exerted by a surface perpendicular to an object resting on it.
On an inclined plane, this force counteracts the component of gravitational force that presses an object into the surface.

• \[ N = mg \cos \theta \]

This equation describes the normal force on an incline, where \( \theta \) is the angle of the plane.
Since normal force is crucial for calculating the static friction force, understanding it gives insight into how it helps balance forces at work.
In the context of the inclined bed, the normal force is integral to maintaining the patient’s position as it influences the static friction that resists their sliding.