**Example 5.6: Acceleration Down a Hill**Now we’re ready to tackle the problem of an object sliding down a frictionless inclined plane. Suppose a child’s toboggan with negligible weight slides down a frictionless hill inclined at an angle α. The toboggan starts from rest. Find the acceleration of the toboggan. Let’s call the direction up the hill the positive x-axis and perpendicular to the hill's surface the positive y-axis.**SET UP**Figure 5.8 is our sketch for this problem. The only force acting on the toboggan is gravity (the weight, **w**). The component of the weight parallel to the hill causes the toboggan to accelerate. The perpendicular component of weight is balanced by the normal force.**SOLUTION**There is only one x-component of force, so\[ \sum F_x = w \sin \alpha = ma_x \]and since \( w = mg \), the acceleration is\[ a_x = g \sin \alpha \]The component equation \(\sum F_y = 0\) gives \( N - w \cos \alpha = 0 \). We see that the component of acceleration is zero because there is no net y-force.**REFLECT**The situations for different angles appear in this expression of \( a_x = g \sin \alpha \). When \( \alpha = 0 \), the acceleration is zero, as we should expect. This means any toboggan “sliding” down a hill will have no y-component of acceleration. When N equals w, the toboggan is stationary. A steep hill (α = 90°) leads to maximum acceleration: \( a_x = g \).**Figure 5.8**(a) The situation described.(b) A free-body diagram shows the forces: **w** is decomposed into two components, one parallel \( (w \sin \alpha) \), and one perpendicular \( (w \cos \alpha) \). **Practice Problem**What angle does the hill slope if the acceleration is 9/2? Answer: 30°. **Page 128 Practice Problem 5.6:**At what angle does the hill slope if the acceleration is \( g/2 \)? **Answer:** \( 30^\circ \)

The case of downward motion on a friction-less inclined plane is given and it is required to find out the inclination angle α such that the acceleration of the mass is g2. Analysis: To find the inclination angle, the Newtons law of motion is to be used along and perpendicular to the motion. There is no motion perpendicular to the plane, first law is used in this direction. There is an acceleration a2 needed in the the direction along the plane, the second law is used in direction along the plane.Apply Newton's first law perpendicular to the plane- n = W cosα Apply newton's second law along the motion of the mass- W sinα = mamg sinα = mg2 sinα =12 α = 300Conclusion: The desired angle of inclination is 300.

Answered: Page 128 Practice Problem 5.6: At what…

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Page 128 Practice Problem 5.6: At what angle does the hill slope if the acceleration is g/2? Answer: 30°.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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