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A \(1125 \mathrm{~kg}\) car and a \(2250 \mathrm{~kg}\) pickup truck approach a curve on a highway that has a radius of \(225 \mathrm{~m}\). (a) At what angle should the highway engineer bank this curve so that vehicles traveling at \(65.0 \mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at \(65.0 \mathrm{mi} / \mathrm{h},\) find the normal force on each one due to the highway surface.

Short Answer

Expert verified
The highway should be banked at an angle of approximately 17 degrees for vehicles traveling at a speed of 65 mi/hr. The heavy truck does not necessarily have to go slower than the lighter car. The normal force on the car and truck are 12,796 N and 25,592 N respectively.

Step by step solution

01

Convert Speed To SI Units

The first step in this process is to convert the speed from miles per hour to the SI unit of meters per second. We use the conversion factor \(1\ mi = 1609\ m\) and \(1\ hr = 3600\ sec\). So, \(v = 65\ mi/hr * (1609\ m/1\ mi) * (1\ hr/3600\ sec) = 29.06\ m/sec\).
02

Calculate the Banking Angle That Allows Safe Turning

The banking angle, θ, that allows a car to turn safely is given by the equation \(\theta = arctan(\frac{v^2}{rg})\) where \(v\) is the speed of the vehicles, \(r\) is the radius of the curve, and \(g\) is the acceleration due to gravity. Substituting the values into the equation gives \(\theta = arctan(\frac{(29.06\ m/s)^2}{(225\ m)(9.8\ m/s^2)}) = 16.5 \supcirc \approx 17\ \supcirc \). This means that the highway should be banked at an angle of around 17 degrees for vehicles traveling at a speed of 65 mi/hr.
03

Discuss If the Heavy Truck Should Go Slower Than the Lighter Car

From physics, we know that both light and heavy vehicles can safely travel at the same speed around a banked curve, given that the curve was designed taking into account that particular speed and the consequences of the object's weight are accounted in the normal force. Hence, the heavy truck should not necessarily go slower than the lighter car.
04

Calculate the Normal Force on Each Vehicle

The normal force on a vehicle moving in a circle on a banked road is given by \(N = \frac{m * v^2}{r * cos\theta}+ m*g* sin\theta \). For the car, \(N_{car} = \frac{(1125\ kg * (29.06\ m/s)^2}{225\ m * cos16.5} + (1125\ kg*9.8\ m/s^2) * sin16.5) = 12796\ N\). For the truck, \(N_{truck}= \frac{(2250\ kg * (29.06\ m/s)^2}{225\ m * cos16.5} + 2250\ kg*9.8\ m/s^2 * sin 16.5)=25592\ N\). This indicates that the normal force on the truck is greater than that on the car due to it being heavier.

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Key Concepts

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

Normal Force
Normal force is a crucial concept in physics, particularly when analyzing how objects move on inclined surfaces like banked curves. It is the perpendicular force exerted by a surface against an object resting on it. In the context of a vehicle on a banked curve, the normal force is significantly influenced by the vehicle's mass and its motion through the curve.

The formula for normal force on a banked curve takes into account both the gravitational force acting on the vehicle and the force required to keep the vehicle in circular motion. It is given by:
  • Normal force, \( N \), can be represented as: \[N = \frac{m \cdot v^2}{r \cdot \cos\theta} + m \cdot g \cdot \sin\theta\]
  • Where:
    • \( m \) is the mass of the vehicle,
    • \( v \) is the velocity,
    • \( r \) is the radius of the curve,
    • \( \theta \) is the banking angle,
    • and \( g \) is the acceleration due to gravity.
For example, in the given scenario, with known values of mass and speed for a car and truck, calculating the normal force helps in understanding the forces at play and ensures the vehicle's stability on the banked curve.
Vehicle Dynamics
Vehicle dynamics is a broad field that examines the forces acting on a moving vehicle and how these forces affect its motion. When navigating a banked curve, the dynamics become quite intriguing.

In a banked curve, the direction and magnitude of forces like gravitational force, friction, and normal force determine how safely a vehicle can round the curve. The balance among these forces ensures vehicles maintain a stable path without skidding. The following points are crucial:
  • All vehicles experience a centripetal force keeping them on the curved path, directed towards the center of the curve.
  • The normal force and gravitational force combine to provide the required centripetal force at specific bank angles.
  • Proper banking angles and vehicle speeds help in minimizing dependence on friction for safe curve navigation.
Understanding these dynamics aids highway engineers in designing safe roads that accommodate vehicles of different weights without requiring them to adjust speed unnecessarily.
Banking Angle Calculation
The banking angle in a curve is calculated to ensure that vehicles can safely navigate the curve at a predetermined speed without relying on friction. This calculation becomes essential in maintaining safety for all vehicles, regardless of their mass.

The formula to calculate the banking angle \( \theta \) is derived from the balance of forces acting on a vehicle in circular motion, given by:
  • The equation is: \[\theta = \arctan\left(\frac{v^2}{r \cdot g}\right)\]
  • Where:
    • \( v \) is the vehicle speed,
    • \( r \) is the radius of the curve,
    • and \( g \) is the acceleration due to gravity.
In our scenario, using the values provided, the highway engineer calculates a banking angle of approximately 17 degrees. Such an angle ensures that vehicles traveling at 65 mph can round the curve safely. Additionally, the calculation indicates there is no need for a heavier vehicle like a truck to slow down compared to a lighter vehicle, such as a car, thereby reinforcing the principle of vehicle dynamics on banked curves.

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Most popular questions from this chapter

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.64). (a) Draw a free-body diagram of the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

A flat (unbanked) curve on a highway has a radius of \(170.0 \mathrm{~m}\). A car rounds the curve at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one-third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

A box with mass \(10.0 \mathrm{~kg}\) moves on a ramp that is inclined at an angle of \(55.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_{\mathrm{k}}=0.300 .\) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F=120.0 \mathrm{~N}\) that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

\(\mathrm{A}\) small car with mass \(0.800 \mathrm{~kg}\) travels at constant speed on the inside of a track that is a vertical circle with radius \(5.00 \mathrm{~m}\) (Fig. E5.45). If the normal force exerted by the track on the car when it is at the top of the track (point \(B\) ) is \(6.00 \mathrm{~N}\), what is the normal force on the car when it is at the bottom of the track (point \(A\) )?

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

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