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Chapter 10: Problem 12

To measure the static friction coefficient between a \(1.6-\mathrm{kg}\) block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant \(=510 \mathrm{N} / \mathrm{m}\) ) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by \(0.039 \mathrm{m} .\) What is the coefficient of static friction?

Short Answer

Expert verified
The coefficient of static friction is approximately 0.789.

Step by step solution

01

Identify Forces and Given Values

The block is held against the vertical wall with the spring, preventing it from slipping down. We are given:- Mass of block, \( m = 1.6 \text{ kg} \).- Spring constant, \( k = 510 \text{ N/m} \).- Spring compression, \( x = 0.039 \text{ m} \).- Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \).We need to find the coefficient of static friction \( \mu_s \).
02

Calculate the Force Exerted by the Spring

The force exerted by the spring when compressed is given by Hooke's law, \( F_{\text{spring}} = kx \).Substitute the given values:\[ F_{\text{spring}} = 510 \text{ N/m} \times 0.039 \text{ m} = 19.89 \text{ N} \]
03

Calculate the Gravitational Force on the Block

The gravitational force on the block is given by \( F_{\text{gravity}} = mg \).Substitute the known values:\[ F_{\text{gravity}} = 1.6 \text{ kg} \times 9.8 \text{ m/s}^2 = 15.68 \text{ N} \]
04

Analyze the Forces

The static friction force must balance the gravitational force to prevent the block from slipping, so \( F_{\text{friction}} = F_{\text{gravity}} \).Since static friction also equals \( \mu_s \times F_{\text{normal}} \), where \( F_{\text{normal}} \) is the force exerted by the spring, the equation becomes:\[ \mu_s \times F_{\text{spring}} = F_{\text{gravity}} \] Substitute values:\[ \mu_s \times 19.89 \text{ N} = 15.68 \text{ N} \]
05

Solve for the Coefficient of Static Friction

Rearrange the equation from Step 4 to solve for \( \mu_s \):\[ \mu_s = \frac{F_{\text{gravity}}}{F_{\text{spring}}} \]Substitute the values:\[ \mu_s = \frac{15.68 \text{ N}}{19.89 \text{ N}} \approx 0.789 \]

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Key Concepts

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

Understanding Static Friction
Static friction is the force that keeps an object at rest when it's placed on a surface. It acts against any attempted motion, preventing the object from sliding or slipping. The magnitude of static friction depends on two main factors: the nature of the surfaces in contact and the normal force pressing them together.
  • The "normal force" is essentially the perpendicular force acting between a surface and an object.
  • An increase in the normal force increases the static frictional force.
Static friction is crucial in preventing items from sliding off surfaces, such as books on a slanted bookshelf or a car parked on a hill. In our exercise, the static friction force must be strong enough to balance out gravitational pull on the block, which asserts downward pressure and would cause it to slide. This particular physics problem demonstrates how static friction can be calculated using the coefficient of friction, which represents the "stickiness" between two surfaces.
Hooke's Law and Spring Force
Hooke's Law offers a straightforward way to describe how springs stretch or compress. When a spring is either extended or compressed, this law precisely quantifies the force exerted by the spring:\[ F_{\text{spring}} = k \times x \]
  • "Fspring" stands for the force the spring exerts.
  • "k" is the spring constant, which tells us about the stiffness of the spring.
  • "x" is the distance by which the spring is compressed or extended.
In our exercise, the spring was compressed by 0.039 meters. Given the spring constant of 510 N/m, Hooke's Law helps calculate the force exerted by the spring to keep the block from falling. This force directly impacts the normal force, thereby influencing the static friction necessary to balance the gravitational pull.
The Role of Gravitational Force
Gravitational force is what pulls objects downwards towards the center of the Earth. For most everyday objects on Earth, gravity imparts a constant acceleration of approximately 9.8 m/s虏. This force is a product of the object's mass and gravitational acceleration, given by the formula:\[ F_{\text{gravity}} = m \times g \]
  • "Fgravity" represents the gravitational force.
  • "m" is the mass of the object in kilograms.
  • "g" is the acceleration due to gravity, usually 9.8 m/s虏 on Earth's surface.
In the physics problem under discussion, the gravitational force pulls the block downward, necessitating counteraction by the static friction to maintain equilibrium. In this situation, the gravitational pull is calculated to ensure it matches the static friction force, resulting in no movement. This emphasis on balancing forces is key to understanding and correctly solving the problem at hand.

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Most popular questions from this chapter

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

A spring stretches by \(0.018 \mathrm{m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m}\). The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A \(1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a 2.50 -kg wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of \(0.200 \mathrm{m} .\) What is the speed of the bullet?
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