/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 ssM A car weighing \(10.7 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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ssM A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m} .(\mathrm{a})\) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is \(0.350\), is the attempt at taking the curve successful?

Short Answer

Expert verified
(a) 3202.68 N of friction is required. (b) Yes, the car can successfully take the curve.

Step by step solution

01

Identify Given Information

We have the following data: - Car weight: \(10.7 \text{ kN}\) (which is equivalent to \(10,700 \text{ N}\)),- Velocity of the car: \(13.4 \text{ m/s}\),- Radius of the curve: \(61.0 \text{ m}\),- Coefficient of static friction: \(0.350\).
02

Calculate the Necessary Frictional Force

To maintain the car on its circular path, the centripetal force required can be found by using the formula: \[ F_c = \frac{mv^2}{r} \] where \(m\) is the mass of the car, \(v\) is the velocity, and \(r\) is the radius.First, calculate the mass of the car: \[ m = \frac{\text{weight}}{g} = \frac{10,700 \text{ N}}{9.8 \text{ m/s}^2} \approx 1092.86 \text{ kg} \].Now, substitute these values into the centripetal force formula:\[ F_c = \frac{(1092.86 \text{ kg}) \times (13.4 \text{ m/s})^2}{61.0 \text{ m}} = 3202.68 \text{ N} \]
03

Check if Static Friction is Sufficient

The maximum static frictional force \(F_s\) that can act is given by:\[ F_s = \mu_s \times N \]where \( \mu_s = 0.350 \) and \( N = \text{weight of the car} = 10,700 \text{ N}\).Therefore, \[ F_s = 0.350 \times 10,700 \text{ N} = 3745 \text{ N} \].
04

Compare Required Friction to Maximum Static Friction

Compare the required frictional force for circular motion \( F_c = 3202.68 \text{ N} \) with the maximum static frictional force \( F_s = 3745 \text{ N} \).Since \( F_s > F_c \), the static friction is sufficient to keep the car on its path.

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Key Concepts

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

Frictional Force
Whenever an object moves or tries to move across a surface, it experiences frictional force. This force opposes the motion, acting in the opposite direction. Frictional force is essential for many everyday activities, like walking or driving, as it prevents slipping. When a car moves along a road, it's the frictional force between the tires and the surface that keeps it from sliding off the road.

There are different types of friction: kinetic, which acts on moving objects, and static, which acts between objects that aren't moving relative to each other. In the context of a car taking a circular turn, it’s the static frictional force that primarily provides the necessary grip to avoid skidding.
  • Friction is dependent on the surfaces in contact.
  • The amount of frictional force is determined by the nature of the materials and how much they are pressed together.
  • Friction is characterized by a friction coefficient, with static friction having its own coefficient, denoted as \( \mu_s \).
Understanding frictional force is crucial in designing vehicles and roads to ensure safety and performance.
Static Friction
Static friction is the force that keeps an object at rest when a force is applied until it overcomes the threshold of motion. For a car on the road, static friction acts between the tires and the road surface preventing the car from slipping sideways during turns.

Unlike kinetic friction, which acts when an object slides over a surface, static friction must be overcome to start motion. This maximum static friction force is called the limiting friction and is calculated by multiplying the coefficient of static friction \( \mu_s \) by the normal force \( N \), which is the weight of the object if on a horizontal surface.
  • The maximum static friction is given as \( F_s = \mu_s \times N \).
  • In circular motion, the static friction provides the required centripetal force to keep the car from deviating from its path.
  • As long as the required centripetal force is less than the maximum static friction, the vehicle remains safely on its path.
By understanding how static friction works, engineers can design safer roads and tires with better grip, optimizing for various driving conditions.
Circular Motion
Circular motion occurs when an object moves along a curved path. For an object to stay in circular motion, a force directed towards the center of the circle—called centripetal force—must act on it. In the case of a car rounding a curve, static friction provides this centripetal force.

Without sufficient centripetal force, an object would move in a straight line due to inertia, according to Newton's first law of motion. For vehicles, this means turning off the intended path. The centripetal force required for circular motion is given by the formula: \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is velocity, and \( r \) is the radius of the circle.
  • Cars need a balance of speed and friction to maintain stable circular motion.
  • Too little friction or excessive speed can cause a loss of control.
  • The radius of the curve and the surface type influence how much frictional force is needed.
Understanding circular motion is vital for traffic safety and vehicle design, helping prevent accidents on curves or during sharp turns.

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Most popular questions from this chapter

Calculate the ratio of the drag force on a jet flying at 1000 \(\mathrm{km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop- driven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km}\). Assume that the air- planes have the same effective cross-sectional area and drag coefficient \(C\).

During an Olympic bobsled run, the Jamaican team makes a turn of radius \(7.6 \mathrm{~m}\) at a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). What is their acceleration in terms of \(g\) ?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the circle's radius is \(10 \mathrm{~m}\). At the top of the circle, what are the (a) magnitude \(F_{B}\) and (b) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?

mechanical engineers must consider how small variations in certain parameters can alter the net force on a passenger. Consider a passenger of mass \(m\) riding around a horizontal circle of radius \(r\) at speed \(v\). What is the variation \(d F\) in the net force magnitude for (a) a variation \(d r\) in the radius with \(v\) held constant, (b) a variation \(d v\) in the speed with \(r\) held constant, and (c) a variation \(d T\) in the period with \(r\) held constant?

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