Problem 16 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at \(28^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30 , respectively. The box slides down unless the applied force has magnitude \(23 \mathrm{~N}\). What is the mass of the box?

Step-by-Step Solution

Answer

The mass of the box is approximately 2.63 kg.

Step 1: Identify Forces Acting on the Box

First, let's identify all the forces acting on the box. The forces include the gravitational force (weight) of the box acting downward, the applied force \( F \) angled at \( 28^{\circ} \) from the horizontal, the normal force from the wall, and the frictional force acting upward opposite to the potential sliding direction.

Step 2: Determine Friction Force Equation

Since the box is on the verge of sliding down, the force of static friction is at its maximum, which counteracts the gravitational pull. The static friction force \( f_s \) is given by \( f_s = \mu_s N \), where \( \mu_s = 0.40 \) is the static friction coefficient and \( N \) is the normal force exerted by the wall.

Step 3: Calculate Components of the Applied Force

Decompose the applied force \( F = 23\, \text{N} \) into its horizontal and vertical components. The horizontal component is \( F_{h} = F \cos(28^{\circ}) \), and the vertical component is \( F_{v} = F \sin(28^{\circ}) \).

Step 4: Determine Normal Force

The normal force \( N \) is equal to the horizontal component of the applied force since they are the only two horizontal forces. Therefore, \( N = F_{h} = 23 \cos(28^{\circ}) \).

Step 5: Formulate the Force Equilibrium Equations

Since the box is in equilibrium vertically (on the verge of moving), the vertical forces must balance. The sum of the frictional force and the vertical component of the applied force equals the weight of the box. Hence the equation is: \( \mu_s N + F_{v} = mg \), where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity \( 9.8\, \text{m/s}^2 \).

Step 6: Solve for Mass of the Box

Using the equation from the previous step, substitute \( F_v = 23 \sin(28^{\circ}) \), \( N = 23 \cos(28^{\circ}) \), and \( \mu_s = 0.40 \) into \( 0.40 \times N + F_{v} = mg \). Then solve for \( m \).

Key Concepts

Force Components

Understanding force components is key when analyzing the movement or stability of objects, especially when forces act at angles. In this exercise, the applied force is at 28° above the horizontal. To work effectively with this force, we break it into two components:

• Horizontal Component (\( F_{h} \)): This part keeps the box pressed against the wall horizontally. It can be calculated using \( F_{h} = F \, \cos(28^{\circ}) \), where \( F \) is 23 N.
• Vertical Component (\( F_{v} \)): This component either aids or counteracts the force of gravity pulling the box down. It is found using \( F_{v} = F \, \sin(28^{\circ}) \).

These components help us understand how the overall force affects the box, and they are crucial for setting up the equilibrium equations. By managing forces in terms of their components, we can easier assess their effects and solve related equations.

Equilibrium of Forces

Equilibrium in physics refers to a state where all forces acting on an object are balanced, resulting in no net force and thus no acceleration. In this scenario, the box is on the verge of sliding down, thus it's at a sort of equilibrium state.

• The box remains at rest due to a balance between the gravitational pull downward and the combined effects of the vertical component of the applied force and static friction.
• Mathematically, this vertical equilibrium is expressed as \( \mu_s N + F_{v} = mg \), where \( \mu_s N \) represents the maximum static friction force, and \( F_{v} = 23 \sin(28^{\circ}) \) is the vertical component of the applied force.
• Horizontally, equilibrium is simpler here since the normal force equals the horizontal component of the applied force, \( N = F_{h} \).

Understanding equilibrium helps us predict whether the object will move or stay put. It involves skillfully setting up force balance equations and solving them to find unknown variables in problems like these.

Normal Force Calculation

The normal force is a supporting force exerted by surfaces against objects resting on them. It's crucial for static friction calculation, which in this problem helps hold the box against sliding down.
For the box holding on a vertical wall, the normal force \( N \) arises from the horizontal component of the applied force, as no other horizontal forces are acting:

• Calculate \( N \) using the horizontal component: \( N = F_h = 23 \cos(28^{\circ}) \).
• This ensures equilibrium horizontally.
• The normal force also influences the maximum static friction, as it directly multiplies with the coefficient of static friction \( \mu_s \).

Correctly determining the normal force is critical since it lets us set the stage for balancing forces vertically and subsequently solving for the box's mass. By grasping how the normal force operates and is calculated, you're one step closer to mastering the friction-related problems.