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Statistics |
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| Total Unique Visitors: 4374393 |
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| Problem 01 | Right Spherical Triangle |
| 2012-01-01 23:16:48 |
Problem
Solve for the spherical triangle whose parts are a = 73°, b = 62°, and C = 90°.
Solution
Encircled the given parts for easy reference
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| Oblique Spherical Triangle |
| 2011-12-10 21:05:16 |
Definition of oblique spherical triangle
Spherical triangles are said to be oblique if none of its included angle is 90° or two or three of its included angles are 90°. Spherical triangle with only one included angle equal to 90° is a right triangle.
Sine law
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| Right Spherical Triangle |
| 2011-12-10 20:12:50 |
Solution of right spherical triangle
With any two quantities given (three quantities if the right angle is counted), any right spherical triangle can be solved by following the Napier’s rules. The rules are aided with the Napier’s circle. In Napier’s circle, the sides and angle of the triangle are written in consecutive order (not including the right angle), and complimentary angles are taken for quantities opposite the right angle.
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| Spherical Trigonometry |
| 2011-12-10 18:54:10 |
Spherical Triangle
Any section made by a cutting plane that passes through a sphere is circle. A great circle is formed when the cutting plane passes through the center of the sphere. Spherical triangle is a triangle bounded by arc of great circles of a sphere.
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| Problem 544 | Friction on Wedges |
| 2011-11-23 00:40:58 |
Problem 544
The block A in Fig. P-544 supports a load W and is to be raised by forcing the wedge B under it. If the angle of friction is 10° at all surfaces in contact, determine the maximum wedge angle α that will give the wedge a mechanical advantage; i.e., make P less than the weight W of the block.
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| Problem 543 | Friction on Wedges |
| 2011-11-22 08:50:10 |
Problem 543
To adjust the vertical position of a column supporting 200-kN load, two 5° wedges are used as shown in Fig. P-543. Determine the force P necessary to start the wedges is the angle of friction at all contact surfaces is 25°. Neglect friction at the rollers.
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| Problem 542 | Friction on Wedges |
| 2011-11-22 08:35:12 |
Problem 542
What force P must be applied to the wedges shown in Fig. P-542 to start them under the block? The angle of friction for all contact surfaces is 10°.
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| Problem 541 | Friction on Wedges |
| 2011-11-22 08:21:57 |
Problem 541
Determine the force P required to start the wedge shown in Fig. P-541. The angle of friction for all surfaces in contact is 15°.
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| Problem 540 | Friction on Wedges |
| 2011-11-22 01:14:57 |
Problem 540
As shown in Fig. P-540, two blocks each weighing 20 kN and resting on a horizontal surface, are to be pushed apart by a 30° wedge. The angle of friction is 15° for all contact surfaces. What value of P is required to start movement of the blocks? How would this answer be changed if the weight of one of the blocks were increased by 30 kN?
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| Problem 539 | Friction on Wedges |
| 2011-11-22 00:52:43 |
Problem 539
The block A in Fig. P-539 supports a load W = 100 kN and is to be raised by forcing the wedge B under it. The angle of friction for all surfaces in contact is f = 15°. If the wedge had a weight of 40 kN, what value of P would be required (a) to start the wedge under the block and (b) to pull the wedge out from under the block?
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