Question

A box with mass m is dragged across a level floor with coefficient of kinetic friction \(\mu_k\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_k, \theta\), and \(g\), obtain an expression for the magnitude of the 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-kg patient across a floor at constant speed by pulling on him at an angle of 25\(^\circ\) above the horizontal. By dragging 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.

Short answer

(a) \( F = \frac{\mu_k mg}{\cos\theta + \mu_k \sin\theta} \); (b) 292.8 N.

Step-by-step solution

Identify the Forces Acting on the Box

The forces acting on the box include the gravitational force (weight), the normal force, the applied force (at an angle), and the kinetic frictional force. The gravitational force is given by weight: \( F_g = mg \). The applied force can be broken into horizontal \( F_{ax} = F \cos \theta \) and vertical \( F_{ay} = F \sin \theta \) components. Kinetic friction force is \( f_k = \mu_k N \), where \( N \) is the normal force.

Establish Equilibrium Conditions

For the box to move with constant speed, the net horizontal force must be zero, so the applied force \( F_{ax} \) must balance the frictional force \( f_k \). Also, the net vertical force must be zero, which means the normal force \( N \) plus the vertical component of the applied force \( F \sin \theta \) balances the gravitational force \( mg \).

Express Normal Force in Terms of Known Quantities

From the vertical force balance, we have: \( N + F \sin \theta = mg \), simplifying to \( N = mg - F \sin \theta \).

Solve for the Frictional Force

The frictional force is given by \( f_k = \mu_k N \), substituting for \( N \) yields: \( f_k = \mu_k (mg - F \sin \theta) \).

Solve for the Required Force \( F \) Using Horizontal Equilibrium

Horizontal equilibrium implies: \( F \cos \theta = \mu_k (mg - F \sin \theta) \). Rearrange this equation to find \( F \): \[ F = \frac{\mu_k mg}{\cos\theta + \mu_k \sin\theta} \].

Calculate the Force for the Given Scenario

Substitute the given values \( m = 90 \text{ kg}, \mu_k = 0.35, \theta = 25^{\circ}, \text{and } g = 9.8 \text{ m/s}^2 \) into the expression for \( F \):\[ F = \frac{0.35 \times 90 \times 9.8}{\cos(25^{\circ}) + 0.35 \sin(25^{\circ})} \].

Simplify and Calculate Numeric Solution

Calculate \( \cos(25^{\circ}) \approx 0.9063 \) and \( \sin(25^{\circ}) \approx 0.4226 \). Substitute these to simplify: \[ F = \frac{308.7}{0.9063 + 0.35 \times 0.4226} = \frac{308.7}{1.0544} \approx 292.8 \text{ N} \].

Key concepts

Kinetic Friction

Kinetic friction is a type of friction that comes into play when two surfaces slide against each other. It's an opposing force that resists motion. When a box is dragged across the floor, kinetic friction acts opposite to the direction of the box's motion. The strength of this friction is determined by the normal force and the coefficient of kinetic friction, denoted as \(\mu_k\). To calculate kinetic friction, use the formula: \(f_k = \mu_k N\), where \(N\) is the normal force. Kinetic friction does work by dissipating energy, usually in the form of heat, as two surfaces rub against one another. Knowing this, it's clear why understanding kinetic friction is essential in solving problems involving sliding objects.

Force Equilibrium

Force equilibrium occurs when the net force acting on a body is zero. This means all the forces balance each other out, resulting in no acceleration. For the box in the exercise to move at a constant speed, it should be in horizontal force equilibrium.

• Horizontal equilibrium: The horizontal component of the applied force should equal the kinetic friction.
• Vertical equilibrium: The normal force plus the vertical component of the applied force should equal the gravitational force.

When these conditions are satisfied, the box moves smoothly without changing its speed. A thorough grasp of force equilibrium helps to ensure systems function in a stable manner without unpredictable movements.

Normal Force

The normal force is the supportive force exerted by a surface perpendicular to the object resting on it. It balances the gravitational force acting on the object. In our case, the box is subject to such a force from the floor. To find the expression for the normal force, consider the vertical forces:\( N + F \sin \theta = mg \), where \(N\) is the normal force, \(F \sin \theta\) is the vertical component of the pulling force, and \(mg\) is the gravitational force on the box.Solving this gives: \N = mg - F \sin \theta\. This adjustment to the normal force due to the vertical component of the pulling force is crucial as it alters the kinetic friction, impacting how easily the box slides.

Horizontal Force Component

When a force acts on an object at an angle, it has both vertical and horizontal components. The horizontal component aids in moving the object across a surface. In our exercise, the horizontal force component is calculated using the expression \(F_{ax} = F \cos \theta\).

• This force must overcome the kinetic friction for the box to move at a constant speed.
• To ensure equilibrium, \(F \cos \theta\) must equal the frictional force \(f_k = \mu_k N\).

By calculating this component accurately, we determine the force required to move objects efficiently along surfaces while combating the resisting frictional forces.