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Chapter 4: Problem 41

(II) 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 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 are given the mass of a silverware drawer and the force needed to overcome static friction and open the drawer. We need to find the coefficient of static friction. The mass of the drawer is 2.0 kg, and the force required to open it is 9.0 N.
02

Define the Static Friction Force

The static friction force is the force that must be overcome to start moving the drawer. It is given by the maximum force that can be applied before the drawer starts to move, which is 9.0 N in this case.
03

Calculate the Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it. For the drawer, the normal force is equal to its weight because it is resting horizontally. The weight is calculated by multiplying the mass with gravity: \[ F_N = mg \]where \( m = 2.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Therefore, \[ F_N = 2.0 \times 9.8 = 19.6 \text{ N} \].
04

Use the Formula for Static Friction Coefficient

The coefficient of static friction \( \mu_s \) is the ratio of the force of static friction to the normal force. It is given by the formula:\[ \mu_s = \frac{F_s}{F_N} \]where \( F_s = 9.0 \text{ N} \).
05

Calculate the Coefficient of Static Friction

Substitute the known values into the equation for \( \mu_s \):\[ \mu_s = \frac{9.0}{19.6} \approx 0.459 \].

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Key Concepts

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

Static Friction
Static friction acts between two surfaces that are not moving relative to each other. It must be overcome for motion to begin. In the problem with the drawer, the owner increases the pulling force until this threshold is reached where static friction is overcome. Static friction is a force that can vary up to a maximum value before an object begins to slide.
  • In this exercise, the maximum static friction force before the drawer moves is 9.0 N.
  • This force keeps the drawer stationary despite the applied force, up to the point of motion.
Understanding that static friction must be overcome to start motion is essential for solving such problems. It explains why sometimes objects do not move even though they are being pushed or pulled.
Normal Force
Normal force is fundamental in understanding everyday phenomena, such as objects resting on surfaces. This force acts perpendicular to the surface, providing support to the object. It ensures that the object does not move through the surface. In the case of the drawer, the normal force equals the weight of the drawer because there are no additional vertical forces. The weight is a product of the mass of the object and the acceleration due to gravity:
  • Formula: \( F_N = mg \)
  • Given: \( m = 2.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \)
  • Calculation: \( F_N = 19.6 \text{ N} \)
Understanding normal force is crucial when calculating static friction, as it directly influences the force that must be overcome to initiate motion.
Physics Problem Solving
Physics problem solving involves identifying known quantities and applying relevant formulas to find unknowns. To tackle the silverware drawer issue, this systematic approach was used:
  • Identify what was known: the mass and the pulling force.
  • Apply the relevant physics principles, such as weight and normal force calculation.
  • Formulate the problem using the static friction coefficient formula: \( \mu_s = \frac{F_s}{F_N} \).
A clear understanding of each step ensures solutions are logical and correct. It's critical not to skip analyzing basic concepts, as these underpin successful problem solving.
Force and Motion
Force and motion are intertwined concepts in physics. They describe how objects interact and move. Force is a push or pull exerted on an object. Motion is the object's change in position due to applied forces. In the drawer scenario:
  • The force of 9.0 N applied by the owner leads to motion once static friction is overcome.
  • Understanding how forces lead to motion helps in predicting and controlling motion in practical situations.
  • By learning about forces and motion, we can better understand how objects start moving, stop, or change direction.
This knowledge is applicable in many real-world contexts, from opening drawers to moving vehicles.

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Most popular questions from this chapter

(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 N. What is the mass of the box?

(II) Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.

(II) On an icy day, you worry about parking your car in your driveway, which has an incline of 12\(^\circ\). Your neighbor's driveway has an incline of 9.0\(^\circ\), and the driveway across the street is at 6.0\(^\circ\). The coefficient of static friction between tire rubber and ice is 0.15. Which driveway(s) will be safe to park in?

A fisherman in a boat is using a "10-lb test" fishing line. This means that the line can exert a force of 45 N without breaking (1 lb \(=\) 4.45N) (\(a\)) How heavy a fish can the fisherman land if he pulls the fish up vertically at constant speed? (\(b\)) If he accelerates the fish upward at 2.0 m/s\(^2\) what maximum weight fish can he land? (\(c\)) Is it possible to land a 15-lb trout on 10-lb test line? Why or why not?

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 \(g\) of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
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