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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 square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2}\) kg. The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91 . A force is applied to the top of the plate, as in Figure \(10.29 a .\) Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{m} .\) The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops \(0.150 \mathrm{m}\) before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

A person who weighs \(670 \mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{cm} .\) (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)
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