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Practice problem: what ramp angle would be needed in order for the cable tension to equal one-half of the car’s weight? Answer 30 degrees

Practice problem: what ramp angle would be needed in order for the cable tension to equal one-half of the car’s weight? Answer 30 degrees

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter6: Applications Of Newton's Laws
Section: Chapter Questions
Problem 41P: A roller coaster car starts from rest at the top of a track 30.0 m long and inclined at 20.0to the...
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Practice problem: what ramp angle would be needed in order for the cable tension to equal one-half of the car’s weight? Answer 30 degrees
ution ΣF = 0, Ty sin 60° + (-₁) = 0. Because T₁=w, we can rewrite the second equation as T₁ T₁= sin 60° W ► FIGURE 5.3 sin 60° <= 1.155 w. EXAMPLE 5.3 Car on a ramp In this example we will see how to handle the case of an object resting on an inclined plane. A car with a weight of 1.76 x 104 N rests on the ramp of a trailer (Figure 5.3a). The car's brakes and transmission lock are re- leased; only the cable prevents the car from rolling backward off the trailer. The ramp makes an angle of 26.0° with the horizontal. Find the tension in the cable and the force with which the ramp pushes on the car's tires. Ta W (a) Car on ramp weight of the engine. If this seems strange, observe that bege enough for its vertical component to be equal in magnitude to w, so 7, itself must have a magnitude larger than w. SOLUTION SET UP Figure 5.3b shows our free-body diagram for the car. The three forces exerted on the car are its weight (magnitude w), the ten- sion in the cable (magnitude 7), and the normal force with magnitude n exerted by the ramp. (Because we treat the car as a particle, we can lump the normal forces on the four wheels together as a single force.) Practice Problem: If we change the angle of the upper chain from 60° to 45°, determine the new expressions for the three tensions. Answers: T₁= w. T₂ = w cos 45/sin 45°=w, Ty w/sin 45°-V2w. We replace the weight by its components. w sina W cos a V (b) Free-body diagram for car Video Tutor Solution We orient our coordinate axes parallel and perpendicular to the ramp. and we replace the weight force by its components. SOLVE The car is in equilibrium, so we first find the components of each force in our axis system and then apply Newton's first law. To find the components of the weight, we note that the angle a between CONTINUED
Transcribed Image Text:ution ΣF = 0, Ty sin 60° + (-₁) = 0. Because T₁=w, we can rewrite the second equation as T₁ T₁= sin 60° W ► FIGURE 5.3 sin 60° <= 1.155 w. EXAMPLE 5.3 Car on a ramp In this example we will see how to handle the case of an object resting on an inclined plane. A car with a weight of 1.76 x 104 N rests on the ramp of a trailer (Figure 5.3a). The car's brakes and transmission lock are re- leased; only the cable prevents the car from rolling backward off the trailer. The ramp makes an angle of 26.0° with the horizontal. Find the tension in the cable and the force with which the ramp pushes on the car's tires. Ta W (a) Car on ramp weight of the engine. If this seems strange, observe that bege enough for its vertical component to be equal in magnitude to w, so 7, itself must have a magnitude larger than w. SOLUTION SET UP Figure 5.3b shows our free-body diagram for the car. The three forces exerted on the car are its weight (magnitude w), the ten- sion in the cable (magnitude 7), and the normal force with magnitude n exerted by the ramp. (Because we treat the car as a particle, we can lump the normal forces on the four wheels together as a single force.) Practice Problem: If we change the angle of the upper chain from 60° to 45°, determine the new expressions for the three tensions. Answers: T₁= w. T₂ = w cos 45/sin 45°=w, Ty w/sin 45°-V2w. We replace the weight by its components. w sina W cos a V (b) Free-body diagram for car Video Tutor Solution We orient our coordinate axes parallel and perpendicular to the ramp. and we replace the weight force by its components. SOLVE The car is in equilibrium, so we first find the components of each force in our axis system and then apply Newton's first law. To find the components of the weight, we note that the angle a between CONTINUED
126 CHAPTER 5 Applications of Newton's Laws the ramp and the horizontal is equal to the angle a between the weight Finally, inserting the numerical values w = 1.76 × 10¹ N and a = 26°, we obtain vector and the normal to the ramp, as shown. The angle a is not mea- sured in the usual way, counterclockwise from the +x axis. To find the components of the weight (w, and wy), we use the right triangles in Figure 5.3b. We find that w = -w sina and wy = -w cosa. The equilibrium conditions then give us ΣF, = 0, ΣF, = 0, T + (-w sina) = 0, n+ (-w cosa) = 0. Be sure you understand how the signs are related to our choice of coordinate axis directions. Remember that, by definition, T, w, and n are magnitudes of vectors and are therefore positive. Solving these equations for T and n, we find T = w sina, n = w cos a. T= (1.76 × 104 N) (sin 26°) = 7.72 × 10³ N, n = (1.76 × 104 N) (cos 26°) = 1.58 × 10³ N. REFLECT To check some special cases, note that if the angle a is zero, then sin a = 0 and cos a = 1. In this case, the ramp is horizontal; no cable tension T'is needed to hold the car, and the magnitude of the total normal force n is equal to the car's weight. If the angle is 90° (the ramp is vertical), then sin a = 1 and cos a = 0. In that case, the cable ten- sion Tequals the weight w and the normal force n is zero. We also note that our results would still be correct if the car were on a ramp on a car transport trailer traveling down a straight highway at a constant speed of 65 mi/h. Do you see why? Practice Problem: What ramp angle would be needed in order for the cable tension to equal one-half of the car's weight? Answer: 30°. EXAMPLE 5.4 Lifting granite and dumping dirt Let us again look at an inclined plane problem, but this time one involving two objects that are tied to- gether by a cable. Blocks of granite, each with weight w₁, are being hauled up a 15° slope out of a quarry (Figure 5.4). For environmental reasons, dirt is also being dumped into the quarry to fill up old holes. You have been asked to find a way to use this dirt to move the granite out more easily. You design a system that lets the dirt (weight w, includi 0X30 Video Tutor Solution
Transcribed Image Text:126 CHAPTER 5 Applications of Newton's Laws the ramp and the horizontal is equal to the angle a between the weight Finally, inserting the numerical values w = 1.76 × 10¹ N and a = 26°, we obtain vector and the normal to the ramp, as shown. The angle a is not mea- sured in the usual way, counterclockwise from the +x axis. To find the components of the weight (w, and wy), we use the right triangles in Figure 5.3b. We find that w = -w sina and wy = -w cosa. The equilibrium conditions then give us ΣF, = 0, ΣF, = 0, T + (-w sina) = 0, n+ (-w cosa) = 0. Be sure you understand how the signs are related to our choice of coordinate axis directions. Remember that, by definition, T, w, and n are magnitudes of vectors and are therefore positive. Solving these equations for T and n, we find T = w sina, n = w cos a. T= (1.76 × 104 N) (sin 26°) = 7.72 × 10³ N, n = (1.76 × 104 N) (cos 26°) = 1.58 × 10³ N. REFLECT To check some special cases, note that if the angle a is zero, then sin a = 0 and cos a = 1. In this case, the ramp is horizontal; no cable tension T'is needed to hold the car, and the magnitude of the total normal force n is equal to the car's weight. If the angle is 90° (the ramp is vertical), then sin a = 1 and cos a = 0. In that case, the cable ten- sion Tequals the weight w and the normal force n is zero. We also note that our results would still be correct if the car were on a ramp on a car transport trailer traveling down a straight highway at a constant speed of 65 mi/h. Do you see why? Practice Problem: What ramp angle would be needed in order for the cable tension to equal one-half of the car's weight? Answer: 30°. EXAMPLE 5.4 Lifting granite and dumping dirt Let us again look at an inclined plane problem, but this time one involving two objects that are tied to- gether by a cable. Blocks of granite, each with weight w₁, are being hauled up a 15° slope out of a quarry (Figure 5.4). For environmental reasons, dirt is also being dumped into the quarry to fill up old holes. You have been asked to find a way to use this dirt to move the granite out more easily. You design a system that lets the dirt (weight w, includi 0X30 Video Tutor Solution
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