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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°.

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

College Physics
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter6: Uniform Circular Motion And Gravitation
Section: Chapter Questions
Problem 6CQ: Pace car drivers routinely cut corners as shown in Figure 6.32. Explain how this allows the curve to...
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that compo Teet our answer what is going c on Solution and the sum of the vertical components is EF, T cosB + (-mg). The x component of the acceleration of the car, string, and key is a, a, and the y component of acceleration is zero, so EF, = Tsin ß= max. EF, = T cosß + (-mg) = ma, = 0. EXAMPLE 5.6 Acceleration down a hill Now we are going to look at the problem of an object sliding down a frictionless inclined plane. Suppose a toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle a, and the toboggan is so well waxed that there is virtually no friction. Find the toboggan's acceleration and the magnitude n of the normal force the hill exerts on the toboggan. SOLUTION SET UP Figure 5.8a is our sketch for this problem. The only forc- es acting on the toboggan are its weight w and the normal force n (Figure 5.8b). The direction of the weight is straight downward, but the direction of the normal force is perpendicular to the surface of the hill, at an angle a with the vertical. We take axes parallel and perpendicular to the surface of the hill and resolve the weight into x and y components. SOLVE There is only one x component of force, so ΣF, = w sina. From EF = max, we have w sina = max and since w = mg, the acceleration is ax = g sina. The y component equation gives EF = n + (-mg cosa). We know that the y component of acceleration is zero because there is no motion in the y direction. So EF, = 0 and n = mg cosa. TOUTER! What is the angle B if the acceleration is g/2? Answer: 26.6. REFLECT The mass m does not appear in the expression for a,; this means that any toboggan, regardless of its mass or number of pas- sengers, slides down a frictionless hill with an acceleration of g sina. In particular, when a = 0 (a flat surface with no slope at all), the DO EVEN D&R W COS X Video Tutor Solution ñ w sin x W (a) The situation (b) Free-body diagram for toboggan A FIGURE 5.8 Our diagrams for this problem. acceleration is ax = 0, as we should expect. When the surface is verti- cal, a = 90° and a, = g (free fall). Note that the magnitude n of the normal force exerted on the toboggan by the surface of the hill (n = mg cosa) is proportional to the magnitude mg of the toboggan's weight; the two are not equal, except in the special case where a = 0. Practice Problem: At what angle does the hill slope if the acceleration is g/2? Answer: 30°.
Transcribed Image Text:that compo Teet our answer what is going c on Solution and the sum of the vertical components is EF, T cosB + (-mg). The x component of the acceleration of the car, string, and key is a, a, and the y component of acceleration is zero, so EF, = Tsin ß= max. EF, = T cosß + (-mg) = ma, = 0. EXAMPLE 5.6 Acceleration down a hill Now we are going to look at the problem of an object sliding down a frictionless inclined plane. Suppose a toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle a, and the toboggan is so well waxed that there is virtually no friction. Find the toboggan's acceleration and the magnitude n of the normal force the hill exerts on the toboggan. SOLUTION SET UP Figure 5.8a is our sketch for this problem. The only forc- es acting on the toboggan are its weight w and the normal force n (Figure 5.8b). The direction of the weight is straight downward, but the direction of the normal force is perpendicular to the surface of the hill, at an angle a with the vertical. We take axes parallel and perpendicular to the surface of the hill and resolve the weight into x and y components. SOLVE There is only one x component of force, so ΣF, = w sina. From EF = max, we have w sina = max and since w = mg, the acceleration is ax = g sina. The y component equation gives EF = n + (-mg cosa). We know that the y component of acceleration is zero because there is no motion in the y direction. So EF, = 0 and n = mg cosa. TOUTER! What is the angle B if the acceleration is g/2? Answer: 26.6. REFLECT The mass m does not appear in the expression for a,; this means that any toboggan, regardless of its mass or number of pas- sengers, slides down a frictionless hill with an acceleration of g sina. In particular, when a = 0 (a flat surface with no slope at all), the DO EVEN D&R W COS X Video Tutor Solution ñ w sin x W (a) The situation (b) Free-body diagram for toboggan A FIGURE 5.8 Our diagrams for this problem. acceleration is ax = 0, as we should expect. When the surface is verti- cal, a = 90° and a, = g (free fall). Note that the magnitude n of the normal force exerted on the toboggan by the surface of the hill (n = mg cosa) is proportional to the magnitude mg of the toboggan's weight; the two are not equal, except in the special case where a = 0. Practice Problem: At what angle does the hill slope if the acceleration is g/2? Answer: 30°.
Our free-body diagram is SHOWI the key are its weight w = mg and the string tension 2. We tinction of the acceleration) and the axic to the cal ar 7 When we uvid un atan R Page 128 Practice Problem 5.6: you put At what angle does the hill slope if the acceleration is g/2? Answer: 30°. ON,
Transcribed Image Text:Our free-body diagram is SHOWI the key are its weight w = mg and the string tension 2. We tinction of the acceleration) and the axic to the cal ar 7 When we uvid un atan R Page 128 Practice Problem 5.6: you put At what angle does the hill slope if the acceleration is g/2? Answer: 30°. ON,
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