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Chapter 5: Problem 11

You're on a plane undergoing a banked turn, so steep that out the window you see the ground below. Yet your pretzels stay put on the seatback tray, rather than sliding downward. Why?

Short Answer

Expert verified
The pretzels stay put on the seatback tray during the steep banked turn because the forces of gravity, the normal force from the tray (acting towards the center of the turn), and static friction (which opposes their natural tendency to slide downward under gravity) balance each other. This balance of forces effectively allows the pretzels to undergo a form of circular motion with the tray, keeping them from sliding away.

Step by step solution

01

Understanding Circular Motion and Forces

A key principle in circular motion is that an object moving in a circle at constant speed is experiencing an acceleration toward the center of the circle. This principle is true for the plane undergoing a banked turn. The forces acting on the pretzels, and thus the tray, in this case are \(mg\) directed downward (gravitational force), normal force \(N\) that acts perpendicular to the tray (towards the center of the turn), and static friction force that opposes the natural tendency of the pretzels to slide downward.
02

Analyzing Forces

There are two main forces at play: the gravitational force pulling the pretzels downwards and the normal force of the tray on the pretzels pushing upwards. There is no upward movement of the pretzels, which means the vertical component of the normal force must balance the weight, i.e., \(Nsinθ = mg\), where \(θ\) is the bank angle. The horizontal component of the normal force, \(Ncosθ\), provides the centripetal force required for circular motion. Static friction \(f_s\) aids the normal force in doing this.
03

Balancing the Forces

Under the influence of the normal force and static friction, the pretzels undergo a form of circular motion, which allows them to stay put on the tray. The effects of these forces counteract the force due to gravity and allow the pretzels to stay on the tray during the plane's steep banked turn.

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Most popular questions from this chapter

Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass \(75 \mathrm{kg}\) ) is on a slope at \(12^{\circ}\) to the horizontal, but the lower climber (mass \(63 \mathrm{kg})\) has gone over the edge to a steeper slope at \(38^{\circ},(\mathrm{a})\) Assuming frictionless ice and a massless rope, what's the acceleration of the pair? (b) The upper climber manages to stop the slide with an ice ax. After the climbers have come to a complete stop, what force must the ax exert against the ice?

A police officer investigating an accident estimates that a moving car hit a stationary car at \(25 \mathrm{km} / \mathrm{h}\). If the moving car left skid marks \(47 \mathrm{m}\) long, and if the coefficient of kinetic friction is 0.71 what was the initial speed of the moving car?

A 2.1 -kg mass is connected to a spring with spring constant \(k=150 \mathrm{N} / \mathrm{m}\) and unstretched length \(18 \mathrm{cm} .\) The two are mounted on a frictionless air table, with the free end of the spring attached to a frictionless pivot. The mass is set into circular motion at \(1.4 \mathrm{m} / \mathrm{s}\). Find the radius of its path.

An airplane goes into a turn \(3.6 \mathrm{km}\) in radius. If the banking angle required is \(28^{\circ}\) from the horizontal, what's the plane's speed?

Find an expression for the minimum frictional coefficient needed to keep a car with speed \(v\) on a banked turn of radius \(R\) designed for speed \(v_{0}\)
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