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A 2.0 -kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied force reaches \(9.0 \mathrm{N},\) the drawer suddenly opens, throwing all the utensils to the floor. What is the coefficient of static friction between the drawer and the cabinet?

Short Answer

Expert verified
The coefficient of static friction is approximately 0.459.

Step by step solution

01

Understand the Problem

We need to find the coefficient of static friction between the drawer and the cabinet. This static friction is what prevents the drawer from sliding until a certain force is applied. When the force reaches 9.0 N (the maximum static friction force), the drawer moves.
02

Identify Known Values

We are given the mass of the drawer as 2.0 kg and the force applied before the drawer opens as 9.0 N. We need to use these values to find the coefficient of static friction.
03

Find the Normal Force

The normal force (F_N) is the force exerted by a surface to support the weight of an object resting on it. For the drawer, it equals the gravitational force acting on it, which can be calculated by:\[ F_N = m \cdot g \]where m is the mass (2.0 kg) and g is the acceleration due to gravity (9.8 m/s²). Substituting the given values:\[ F_N = 2.0 \cdot 9.8 = 19.6 \text{ N} \]
04

Use the Formula for Static Friction

The static friction force (F_s) is related to the normal force by the static friction coefficient (\mu_s) as:\[ F_s = \mu_s \cdot F_N \]We know F_s is 9.0 N (the force at which the drawer starts moving). Substituting both F_s and F_N into the formula gives:\[ 9.0 = \mu_s \cdot 19.6 \]
05

Solve for the Coefficient of Static Friction

To find the coefficient of static friction (\mu_s), rearrange the formula:\[ \mu_s = \frac{9.0}{19.6} \approx 0.459 \]

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Key Concepts

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

Coefficient of Static Friction
The coefficient of static friction is a measure that helps us understand how much force is needed to start moving an object at rest. It represents the ratio between the maximum static friction force and the normal force. In this particular problem, the force required to initially move the drawer was 9.0 N, which serves as the maximum static friction force.
To calculate the coefficient of static friction (\( \mu_s \)), you need both the maximum static friction force and the normal force. The relationship is captured by the formula:\[ F_s = \mu_s \cdot F_N \]In this problem, substituting the values gives us:\[ \mu_s = \frac{9.0}{19.6} \approx 0.459 \]This coefficient tells us the interaction level between the two surfaces—in this case, the drawer and the cabinet—shedding light on how much they naturally resist movement against each other. A higher coefficient means more resistance to start moving.
Normal Force
The normal force is essential in physics as it acts perpendicular to the surface support, helping to balance the gravitational pull on an object like our drawer. For the drawer, the normal force (\( F_N \)) equals the gravitational force. This is calculated through the formula:\[ F_N = m \cdot g \]where\( m \) is the object's mass—2.0 kg here—and\( g \) is the gravitational acceleration—9.8 m/s². Thus, the normal force for the drawer is:\[ F_N = 2.0 \times 9.8 = 19.6 \text{ N} \]The normal force helps in determining the static friction force because they are directly proportional. Understanding these forces is critical in physics problem-solving, as they form the basis of analyzing forces acting on stationary objects.
Physics Problem-Solving
Solving physics problems often requires a structured approach that includes understanding the problem, identifying what is known, applying relevant formulas, and solving for the unknown. In our drawer problem:
  • We have identified the need to find the coefficient of static friction.
  • We knew the object's mass and the force applied.
  • We calculated the normal force and utilized the static friction formula.
Breaking problems into these steps can make complex physics challenges more manageable. Consistent practice with different physics scenarios improves problem-solving skills, enabling one to apply the correct formulas and logical thinking effectively to solve problems accurately.
Newton's Laws of Motion
Newton's Laws of Motion offer fundamental insights into how objects behave with forces acting upon them. They play a crucial role in understanding this drawer scenario.
According to Newton's First Law, an object remains at rest or moves at a constant velocity unless acted upon by a net external force. For our drawer, it stayed still due to static friction until the pulling force exceeded this friction.
Newton's Second Law elucidates that force equals mass times acceleration (\( F = m \cdot a \)), helping us understand that the force needed to overcome static friction is indeed linked to the weight of the drawer.
The Third Law, which states that every action has an equal and opposite reaction, makes sense when considering how the drawer exerts equal force back on the hand pulling it once in motion. Understanding these laws helps in dissecting everyday experiences like a stuck drawer, thus connecting the theoretical principles to practical implications.

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Most popular questions from this chapter

(II) \(\mathrm{A}\) 75-kg snowboarder has an initial velocity of 5.0 \(\mathrm{m} / \mathrm{s}\) at the top of a \(28^{\circ}\) incline (Fig. \(36 ) .\) After sliding down the 110 -m long incline (on which the coefficient of kinetic friction is \(\mu_{k}=0.18\) ), the snowboarder has attained a velocity \(v .\) The snowboarder then slides along a flat surface (on which \(\mu_{k}=0.15\) and comes to rest after a distance \(x .\) Use Newton's second law to find the snowboarder's acceleration while on the incline and while on the flat surface. Then use these accelerations to determine \(x .\)

(III) A small block of mass \(m\) rests on the rough, sloping side of a triangular block of mass \(M\) which itself rests on a horizontal frictionless table as shown in Fig. 5-41. If the coefficient of static friction is \(\mu,\) determine the minimum horizontal force \(F\) applied to \(M\) that will cause the small block \(m\) to start moving up the incline.

(II) A \(25.0-\mathrm{kg}\) box is released on a \(27^{\circ}\) incline and accelerates down the incline at \(0.30 \mathrm{~m} / \mathrm{s}^{2}\). Find the friction force impeding its motion. What is the coefficient of kinetic friction?

(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?

(I) A force of \(35.0 \mathrm{~N}\) is required to start a 6.0 -kg box moving across a horizontal concrete floor. ( \(a\) ) What is the coefficient of static friction between the box and the floor? \((b)\) If the 35.0-N force continues, the box accelerates at \(0.60 \mathrm{~m} / \mathrm{s}^{2}\). What is the coefficient of kinetic friction?

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