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Chapter 5: Problem 113

Uterior Motives. You are driving a classic 1954 Nash Ambassador with a friend who is sitting to your right on the passenger side of the front seat. The Ambassador has flat bench seats. You would like to be closer to your friend and decide to use physics to achieve your romantic goal by making a quick turn. (a) Which way (to the left or to the right) should you turn the car to get your friend to slide closer to you? (b) If the coefficient of static friction between your friend and the car seat is \(0.35,\) and you keep driving at a constant speed of 20 \(\mathrm{m} / \mathrm{s}\) , what is the maximum radius you could make your turn and still have your friend slide your way?

Short Answer

Expert verified
(a) Turn right. (b) Maximum radius: 116.6 m.

Step by step solution

01

Understand the Physics of Sliding

To slide your friend towards you, you'll need to make a turn in such a way that your friend slides due to inertia. Inertia will cause objects not firmly fixed to the car to move towards the outer edge of the turning circle. If you turn right, your friend will slide to the left, towards you.
02

Determine the Forces Involved

The force causing your friend to slide is the centripetal force. In this scenario, it is equivalent to the force of friction resisting the sliding motion. The static friction force can be calculated using the formula \( f_{s} = \mu_s \cdot m \cdot g \), where \( \mu_s \) is the static friction coefficient, \( m \) is the mass of the person, and \( g \) is the acceleration due to gravity (9.8 \( \mathrm{m/s^2} \)).
03

Apply Newton's Second Law

The centripetal force needed to keep your friend along the curve of the seat without sliding is given by \( f_{c} = \frac{m \cdot v^2}{r} \). To have your friend slide, this centripetal force must exceed the maximum static friction \( f_{s} \).
04

Calculate Maximum Radius

Set the frictional force equal to the centripetal force and solve for the radius \( r \): \( \mu_s \cdot m \cdot g = \frac{m \cdot v^2}{r} \). Cancel out \( m \) from both sides and solve for \( r \): \[ r = \frac{v^2}{\mu_s \cdot g} \].
05

Substitute Values and Compute

Substitute the given values into the formula: \( v = 20 \; \mathrm{m/s} \), \( \mu_s = 0.35 \), and \( g = 9.8 \; \mathrm{m/s^2} \) to calculate \( r \). \[ r = \frac{20^2}{0.35 \times 9.8} = \frac{400}{3.43} \approx 116.6 \; \mathrm{m} \].

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Key Concepts

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

Static Friction
Static friction is the force that prevents objects from sliding past each other when they are in contact. In this scenario, it is the force that initially keeps your friend from sliding across the car seat. This force arises due to the surface texture and the nature of both the seat material and your friend's clothing.

Static friction can be calculated using the formula:
  • \( f_{s} = \mu_s \cdot m \cdot g \)
where:
  • \( f_{s} \) is the static friction force,
  • \( \mu_s \) is the coefficient of static friction (given as 0.35),
  • \( m \) is the mass of your friend, and
  • \( g \) is the acceleration due to gravity (\( 9.8 \; \mathrm{m/s^2} \)).
This coefficient \( \mu_s \) is crucial. It tells us how much grip there is before sliding occurs. If the centripetal force due to the turn exceeds this static friction, your friend will slide over the seat.

It's important to note that static friction resists motion until a certain threshold is surpassed, which allows sliding to occur.
Inertia
Inertia is the tendency of an object to resist changes in its motion. It is one of the key concepts in understanding why your friend might slide toward you during a turn.

Imagine that when your car turns, everything in the car, including your friend, wants to keep moving straight. This happens because objects naturally want to maintain their state of motion. Therefore, when the car turns right, your friend's body will continue moving straight but the car鈥攁nd you on the left side鈥攑ivot around.
  • This results in your friend sliding leftwards, toward you, due to inertia.
No external force is pushing your friend to the left; it's a result of the change in the car's motion while inertia keeps your friend moving in a straight line.

Understanding inertia is key to predicting how objects behave when subjected to forces.
Newton's Second Law
Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration \( (F = m \cdot a) \). In the context of our car and passenger friend, it helps explain why your friend eventually slides if you turn sharply enough.

When making a turn, the car exerts a centripetal force to keep your friend moving in a circular path. This centripetal force required is defined by the equation:
  • \( f_{c} = \frac{m \cdot v^2}{r} \)
where:
  • \( f_{c} \) is the centripetal force,
  • \( m \) is the mass of your friend,
  • \( v \) is the velocity (20 \( \mathrm{m/s} \)), and
  • \( r \) is the radius of the turn.
Newton's Second Law links these ideas together.

For your friend to slide, this centripetal force must overcome the static friction. By finding the point where static friction equals the required centripetal force, you can calculate the maximum radius of the turn where your friend will slide toward you.

This relationship illustrates the practical application of Newton's assertions in real-life automotive dynamics.

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Most popular questions from this chapter

A box with mass \(m\) is dragged across a level fioor having a coefficient of kinctic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .(\text { a) In }\) terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90-\mathrm{kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the borizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{k}=0.35 .\) Use the result of part (a) to answer the instructor's question.

Wheels. You find that it takes a horizontal force of 160 \(\mathrm{N}\) to slide a box along the surface of a level fioor at constant speed. The coefficient of static friction is \(0.52,\) and the coefficient of kinetic friction is \(0.47 .\) If you place the box on a dolly of mass 5.3 \(\mathrm{kg}\) and with coefficient of rolling friction \(0.018,\) what horizontal acceleration would that \(160-\mathrm{N}\) force provide?

A 8.00 kg block of ice, released from rest at the top of a 1.50 -m-long frictionless ramp, slides downhill, reaching a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 \(\mathrm{N}\) parallel to the surface of the ramp?

A 2540 -kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by \(v(t)=\) \(A t+B t^{2},\) where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and 1.00 s later an upward velocity of 2.00 \(\mathrm{m} / \mathrm{s}\) . (a) Determine \(A\) and \(B,\) including their \(\mathrm{SI}\) units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on of it, assume no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

Losing Cargo. A \(12.0-\mathrm{kg}\) box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{s}=0.19\) and \(\mu_{k}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?
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