91Ó°ÊÓ

A box with mass \(m\) is dragged across a level fioor having a coefficient of kinctic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .(\text { a) In }\) terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90-\mathrm{kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the borizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{k}=0.35 .\) Use the result of part (a) to answer the instructor's question.

Step by step solution

01

Analyzing Forces

To move the box at constant speed, the net force on it must be zero. The main forces acting on the box are the applied force \( F \), gravitational force \( mg \), normal force \( N \), and frictional force \( f_k \). The frictional force can be given by \( f_k = \mu_k N \). The force \( F \) can be split into horizontal \( F \cos(\theta) \) and vertical \( F \sin(\theta) \) components.

02

Calculating Normal Force

The normal force \( N \) is the force exerted by the surface to support the weight of the object. Since the box is dragged at an angle, \( N = mg - F \sin(\theta) \), which accounts for both the gravitational force and the vertical component of the applied force.

03

Friction Force Calculation

Using the relation for frictional force, \( f_k = \mu_k N = \mu_k (mg - F \sin(\theta)) \). This provides the expression for the kinetic friction that opposes the motion.

04

Balancing Horizontal Forces

Since the net force must be zero, the horizontal component of the applied force must equal the frictional force: \( F \cos(\theta) = f_k = \mu_k (mg - F \sin(\theta)) \).

05

Solving for the Required Force

Rearrange the equation from Step 4 to solve for the force \( F \): \[ F \cos(\theta) = \mu_k mg - \mu_k F \sin(\theta) \] \[ F(\cos(\theta) + \mu_k \sin(\theta)) = \mu_k mg \] \[ F = \frac{\mu_k mg}{\cos(\theta) + \mu_k \sin(\theta)} \] This equation gives the magnitude of the force required to move the box at a constant speed.

06

Applying Formula for Specific Values

Now, use the given values to find the force required to drag the patient. The mass \( m = 90 \text{ kg} \), \( \theta = 25^\circ \), and \( \mu_k = 0.35 \) with \( g = 9.8 \text{ m/s}^2 \). Substitute these values into the equation: \[ F = \frac{0.35 \times 90 \times 9.8}{\cos(25^\circ) + 0.35 \sin(25^\circ)} \].

07

Calculating the Exact Force

Calculate the value by resolving the trigonometric functions: \[ F \approx \frac{308.7}{\cos(25^\circ) + 0.35 \sin(25^\circ)} \] Using a calculator to evaluate, we find: \[ F \approx 293.16 \text{ N} \].

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Key Concepts

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

Force Components

When a force is applied at an angle, it can be quite challenging to grasp how it affects an object. Essentially, any force applied to an object can be broken down into two components: horizontal and vertical. These are called force components. The horizontal component is often involved in moving the object across a surface, while the vertical component can affect the normal force (the force perpendicular to the surface).

Using trigonometry, we can express these force components in terms of the applied force and the angle at which it is applied. This is done by the following:

• The horizontal component of the force is determined by multiplying the magnitude of the force by the cosine of the angle: \( F_x = F \cos(\theta) \).
• The vertical component of the force is found by multiplying the force by the sine of the angle: \( F_y = F \sin(\theta) \).

Understanding these components is crucial when analyzing the motion of an object on a plane, as they directly influence other forces, such as the frictional and normal forces.

Normal Force

The normal force is a support force exerted upon an object that is in contact with a stable surface. This force acts perpendicular to the surface. It directly opposes the object's weight when no vertical movement occurs.

In cases where a force is applied at an angle, like pulling a box upward with a rope, the vertical component of the force alters the overall normal force. This effect can be seen with the following equation:

• \( N = mg - F \sin(\theta) \)

This equation illustrates that as the upwards pull (vertical component) increases, it subtracts from the normal force because part of the weight is now being supported by that vertical force instead. It is essential to have a clear understanding of normal force, as it plays a pivotal role in the calculation of frictional force.

Frictional Force

Frictional force acts to resist the motion of an object sliding across a surface. It acts in the opposite direction to movement and is crucial for stopping objects from sliding indefinitely.

The frictional force can be calculated using the equation:

• \( f_k = \mu_k N \).

Here, \( \mu_k \) represents the coefficient of kinetic friction, and \( N \) is the normal force. The coefficient of kinetic friction is a property that depends on the materials in contact and describes the frictional interaction between them.

By knowing the frictional force, we can assess how much force is necessary to maintain constant speed, equating it to the horizontal component of the applied force. This analysis is vital for understanding motion dynamics on a surface.

Trigonometry in Physics

Trigonometry is a powerful tool in physics used to resolve vectors, which are quantities defined by both magnitude and direction. When dealing with forces at angles, trigonometry provides the formulas necessary to determine force components.

In our context, we employ trigonometric functions such as sine and cosine to decompose a force into components that are easier to analyze. For instance, rather than ponder over a single diagonal force vector, decomposing it with

• \( F\sin(\theta) \) representing the vertical influence
• \( F\cos(\theta) \) handling the horizontal influence,

allows us to see exactly how these forces impact the object.

Understanding basic trigonometry is essential in physics, as it allows us to address problems involving forces, such as the one in this exercise, by simplifying complex, angled scenarios into manageable horizontal and vertical influences.