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You are at the shoe store to buy a pair of basketball shoes that have the greatest traction on a specific type of hardwood. To determine the coefficient of static friction. \(\mu\), you place each shoe on a plank of the wood and tilt the plank to an angle \(\theta\), at which the shoe just starts to slide. Obtain an expression for \(\mu\) as a function of \(\theta\).

Short Answer

Expert verified
Answer: The coefficient of static friction as a function of the angle is given by \(\mu = \tan{\theta}\).

Step by step solution

01

Analyze the forces acting on the shoe

When the shoe is placed on the inclined wooden plank, the two forces acting on it are: 1. Gravitational force, which pulls the shoe downwards with a force \(F_g = mg\), where \(m\) is the mass of the shoe, and \(g\) is the acceleration due to gravity. 2. Static friction force, which opposes the gravitational force, preventing the shoe from sliding down the incline. As the plank is tilted, the gravitational force can be decomposed into two components: one parallel to the incline and the other perpendicular to it. The parallel component tries to pull the shoe down along the incline, while the perpendicular component is balanced by the normal force acting between the shoe and the plank.
02

Calculate the gravitational force components

The component of gravitational force parallel to the incline can be calculated as: \(F_{g\parallel} = mg\sin{\theta}\). This is because the angle between the force and the incline is \(\theta\). The component of gravitational force perpendicular to the incline can be calculated as: \(F_{g\perp} = mg\cos{\theta}\), as this component is adjacent to the angle \(\theta\).
03

Find the static friction force

The static friction force (\(F_f\)) is given as \(F_f \leq \mu F_N\), where \(\mu\) is the coefficient of static friction and \(F_N\) is the normal force acting between the shoe and the plank. As the normal force is equal to the gravitational force component perpendicular to the incline (\(F_N = F_{g\perp}\)), we have \(F_f \leq \mu mg\cos{\theta}\).
04

Find the point where the shoe just begins to slide

The shoe will start sliding when the static friction force equals the gravitational force component parallel to the incline (i.e., \(F_f = F_{g\parallel}\)). So, we have \(\mu mg\cos\theta = mg\sin\theta\).
05

Solve for the coefficient of static friction

We want to find an expression for \(\mu\) in terms of \(\theta\). Divide both sides of the equation by \(mg\cos{\theta}\) to find the expression: \(\mu = \dfrac{mg\sin{\theta}}{mg\cos{\theta}}\) Cancel out \(mg\) and simplify the expression: \(\mu = \dfrac{\sin{\theta}}{\cos{\theta}}\) Finally, we arrive at the expression for the coefficient of static friction as a function of \(\theta\): \(\mu = \tan{\theta}\)

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Key Concepts

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

Static Friction
Imagine trying to push a heavy box across the floor. It's tough to get it moving, right? This resistance to start moving is due to static friction. Static friction acts between surfaces at rest relative to each other, preventing motion until a certain threshold force is exceeded. In our exercise, the shoes on the tilted plank experience static friction which holds them steady until the incline becomes too steep and they begin to slide.

This threshold where the shoe just begins to slide indicates the maximum value of static friction. The force of static friction is defined by the equation:
\( F_f = \mu F_N \), where \( \(mu \)\) represents the coefficient of static friction, and \( F_N \) is the normal force. By finding the point where static friction can no longer counteract the gravitational pull down the incline, we manage to deduce the coefficient \( \(mu \)\) as a tangible measure of traction.
Incline Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It's a classic setup to study components of forces and help us understand how objects behave on slopes, like our shoe experiment. When we talk about an inclined plane, we're considering a fundamental physics problem: how does the angle of the incline affect the forces on an object resting on it?

A steeper angle increases the gravitational force component trying to pull the object down, which is also why one feels more pressure walking up a steeper hill. The unique element of our challenge is finding the angle \( \theta \) at which the static friction gives way and the shoe starts to slide down, which is directly correlated with the angle of the incline.
Gravitational Force Components
Gravity is the force that pulls us to the Earth, but when an object is on an incline, this gravitational force isn't just a straight 'downward' pull. It's helpful to break it down into two parts, or components: one that's parallel to the slope (trying to slide the object down the plane) and another that's perpendicular (pushing the object into the slope).

These components are central to our shoe example. They are defined mathematically as:
\( F_{g\parallel} = mg\sin{\theta} \) for the parallel component, and
\( F_{g\perp} = mg\cos{\theta} \) for the perpendicular component. Understanding these components helps us unravel the puzzle of at what angle \( \theta \) the component of gravity along the plane overcomes static friction's grip.
Normal Force
When we stand on the ground, not only are we pulled down by gravity, but we are also pushed up by the ground with a force equal in magnitude and opposite in direction. This support force is called the normal force. It’s 'normal' in the sense that it’s perpendicular to the contact surface.

In the context of our inclined plane situation, the normal force is crucial because it balances out the perpendicular component of the gravitational force. It's represented in equations as:
\( F_N = mg\cos{\theta} \). Importantly, the normal force is tied to the force of static friction, a relationship that forms the core of finding the coefficient of static friction, \( \(mu \)\), for the shoe experiment.

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Most popular questions from this chapter

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

4.71 A block of mass \(20.0 \mathrm{~kg}\) supported by a vertical massless cable is initially at rest. The block is then pulled upward with a constant acceleration of \(2.32 \mathrm{~m} / \mathrm{s}^{2}\). a) What is the tension in the cable? b) What is the net force acting on the mass? c) What is the speed of the block after it has traveled \(2.00 \mathrm{m?}\)

The Tornado is a carnival ride that consists of vertical cylinder that rotates rapidly aboat its vertical axis. As the Tornado rotates, the riders are presscd against the inside wall of the cylinder by the rotation, and the floon of the cylinder drops away. The force that points upward. preventing the riders from falling downward, is a) friction force c) gravity. b) a normal force. d) a tension force.

Which of the following observations about the friction force is (are) incorrect? a) The magnitude of the kinetic friction force is always proportional to the normal force b) The magnitude of the static friction force is always proportional to the normal force. c) The magnitude of the static friction force is always proportional to the external applied force. d) The direction of the kinetic friction force is always opposite the direction of the relative motion of the object writh respect to the surface the object moves on. e) The direction of the static friction force is always opposite that of the impending motion of the object relative to the surface it rests on f) All of the above are correct.

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