Fig. 5.11 in .NET

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Fig. 5.11
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5.6 Inclined Plane
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An object with weight W on an inclined plane which has an angle of inclination exerts a force Fa against the inclined plane and a force Fd down the inclined plane. The forces Fa and Fd are the component vectors for the weight W. See Fig. 5.12(a). The angle formed by the force Fa against the inclined plane and the weight W is equal to the angle of inclination . Since , Fa W cos and Fd W sin . See Fig. 5.12(b). The minimum force needed to keep an object from sliding down an inclined plane (ignoring friction) has the same magnitude but is in the opposite direction from Fd.
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Fig. 5.12
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CHAPTER 5 Practical Applications
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EXAMPLE 5.8 A 500-lb barrel rests on an 11.2 inclined plane. What is the minimum force (ignoring friction) needed to keep the barrel from rolling down the incline and what is the force the barrel exerts against the surface of the inclined plane (See Fig. 5.13.)
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Fig. 5.13
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Fd Fa
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Manual Fa Fa 500(0.9810) 490.5 491 lb
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500 sin 11.2 500(0.1942) 97.1 97.1 lb 500 cos 11.2
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Calculator Fa Fa 500(0.980955) 490.478 490 lb
The minimum force needed to keep the barrel from rolling down the incline is 97.1 lb and the force against the inclined plane is 491 lb (or 490 lb using a calculator).
SOLVED PROBLEMS
Use the rounding procedures stated in Sec. 4.7. 5.1 A motorboat moves in the direction N40 E for 3 h at 20 mi/h. How far north and how far east does it travel
Suppose the boat leaves A. Using the north-south line through A, draw the ray AD so that the bearing of D from S A is N40 E. On AD locate B such that AB 3(20) 60 mi. Through B, pass a line perpendicular to the line NAS, meeting it in C. In the right triangle ABC, (see Fig. 5.14), AC and CB AB cos A AB sin A 60 cos 40 60 sin 40 60(0.7660) 60(0.6428) 45.96 38.57
The boat travels 46 mi north and 39 mi east.
Fig. 5.14
CHAPTER 5 Practical Applications
5.2 Three ships are situated as follows: A is 225 mi due north of C, and B is 375 mi due east of C. What is the bearing of (a) of B from A and (b) of A from B
In the right triangle ABC, see Fig. 5.15, tan / CAB 375/225 1.6667 and / CAB 59 0
(a) The bearing of B from A (angle SAB) is S59 0 E. (b) The bearing of A from B (angle N BA) is N59 0 W.
Fig. 5.15
5.3 Three ships are situated as follows: A is 225 miles west of C while B, due south of C, bears S25 10 E from A. (a) How far is B from A (b) How far is B from C (c) What is the bearing of A from B
From Fig. 5.16, / SAB AB or and (c) Since / CBA AB CB 25 10 and / BAC AC sec / BAC AC/cos / BAC AC tan / BAC 64 50 . Then 225(2.3515) 225/0.4253 225(2.1283) 529.1 529.0 478.9 225 sec 64 50 225/cos 64 50 225 tan 64 50
(a) B is 529 miles from A.
(b) B is 479 miles from C.
25 10 , the bearing of A from B is N25 10 W.
Fig. 5.16
5.4 From a boat sailing due north at 16.5 km/h, a wrecked ship K and an observation tower T are observed in a line due east. One hour later the wrecked ship and the tower have bearings S34 40 E and S65 10 E. Find the distance between the wrecked ship and the tower.
In Fig. 5.17, C, K, and T represent, respectively, the boat, the wrecked ship, and the tower when in a line. One hour later the boat is at A, 16.5 km due north of C. In the right triangle ACK, CK 16.5 tan 34 40 16.5(0.6916)
CHAPTER 5 Practical Applications
In the right triangle ACT, CT Then KT 16.5 tan 65 10 CT CK 16.5(2.1609) 0.6916) 24.2 km
16.5(2.1609
Fig. 5.17
5.5 A ship is sailing due east when a light is observed bearing N62 10 E. After the ship has traveled 2250 m, the light bears N48 25 E. If the course is continued, how close will the ship approach the light
In Fig. 5.18, L is the position of the light, A is the first position of the ship, B is the second position, and C is the position when nearest L. In the right triangle ACL, AC CL cot / CAL CL cot 27 50 1.8940CL. In the right triangle BCL, BC Since AC BC CL cot / CBL 1.1270CL CL cot 41 35 2250, 1.8940CL 1.1270CL. 2250 2250 and CL 1.8940 1.1270 2934 m.
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