Solutions

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 2 2 + 2 2 + 3 2 = v1 + v2 + v3 / v v2 v v 2 = v 2/ v 2 =1 18. Let v = x, y, z , then x = x2 + y 2 cos θ, y = x2 + y 2 sin θ, x2 + y 2 = v cos λ, and z = v sin λ, so x/ v = cos θ cos λ, y/ v = sin θ cos λ, and z/ v = sin λ. √ √√ √ 3 1 31 3 33 = , cos β = = , cos γ = ; α ≈ 64◦ , β ≈ 41◦ , γ = 60◦ 19. cos α = 22 4 22 4 2 20. Let u1 = u1 cos α1 , cos β1 , cos γ1 , u2 = u2 cos α2 , cos β2 , cos γ2 , u1 and u2 are perpendicular if and only if u1 · u2 = 0 so u1 u2 (cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 ) = 0, cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 = 0. 21. (a) b = 3/5, 4/5 , so projb v = 6/25, 8/25 b and v − projb v = 44/25, −33/25 y 2 projbv x 2 v -2 -2 (b) √ √ b = 1/ 5, −2/ 5 , so projb v = −6/5, 12/5 b and v − projb v = 26/5, 13/5 5 v – projbv y v – projbv v projbv -5 x 5 -5 Exercise Set 13.3 (c) 460 √ √ b = 2/ 5, 1/ 5 , so projb v = −16/5, −8/5 b and v − projb v = 1/5, −2/5 y x -4 v – projbv projbv v -4 22. (a) (b) b = 1/3, 2/3, 2/3 , so projb v = 2/3, 4/3, 4/3 and v − projb v = 4/3, −7/3, 5/3 b b = 2/7, 3/7, −6/7 , so projb v = −74/49, −111/49, 222/49 b and v − projb v = 270/49, 62/49, 121/49 23. (a) projb v = −1, −1 , so v = −1, −1 + 3, −3 (b) projb v = 16/5, 0, −8/5 , so v = 16/5, 0, −8/5 + −1/5, 1, −2/5 24. (a) projb v = 1, 1 , so v = 1, 1 + −4, 4 (b) projb v = 0, −8/5, 4/5 , so v = 0, −8/5, 4/5 + −2, 13/5, 26/5 −→ −→ −→ 25. AP = −i + 3j, AB = 3i + 4j, proj −→ AP AB −→ √ AP = 10, 10 − 81/25 = 13/5 −→ −→ −→ −→ −→ = | AP · AB |/ AB −→ 26. AP = −4i + 2k, AB = −3i + 2j − 4k, proj −→ AP AB −→ √ AP = 20, 20 − 16/29 = 564/29 = 9/5 −→ −→ −→ √ = | AP · AB |/ AB = 4/ 29. 27. Let F be the downward force of gravity on the block, then F = 10(9.8) = 98 N, and if F1 and √ F2 are the forces parallel to and perpendicular to the ramp, then F1 = F2 = 49 2 N. Thus √ √ the block exerts a force of 49 2 N against the ramp and it requires a force of 49 2 N to prevent the block from sliding down the ramp. 28. Let x denote the magnitude of the force in the direction of Q. Then the force F acting on the 1 1 block is F = xi − 10j. Let u = − √ (i + j) and v = √ (i − j) be the unit vectors in the directions 2 2 x + 10 x − 10 along and against the ramp. Then F decomposes as F = − √ u + √ v, and thus the block 2 2 will not slide down the ramp provided x ≥ 10 N. 29. Three forces act on the block: its weight −300j; the tension √ cable A, which has the form in j a(−i + j); and the tension in cable B, which has the form b( 3i −√), where a, b are positive √ constants. The sum of these forces is zero, which yields a = 450 + 150 3, b = 150 + 150 3. Thus the forces√ along cables A and B are, √ respectively, √ √ √ √ 150(3 + 3)(i − j) = 450 2 + 150 6 lb, and 150( 3 + 1)( 3i − j) = 300 + 300 3 lb. 461 Chapter 13 30. (a) Let T and T be the forces exerted on the block by cables A and B . Then A B T = a(−10i + dj) and T = b(20i + dj) for some positive a, b. Since T + T − 100j = 0, we A B A B 100 200 100 2000 2000 200 ,b = ,T = − i+ j, and T = i+ j. find a = A B 3d 3d 3d 3 3d 3 500 -20 100 -100 (b) An increase in d will decrease both forces. 500 -20 100 -100 40 ≤ 150 is equivalent to d ≥ √ , and T B 65 40 40 . d ≥ √ . Hence we must have d ≥ 65 77 (c) The inequality T A −→ ≤ 150 is equivalent to −→ 31. Let P and Q be the points (1,3) and (4,7) then P Q = 3i + 4j so W = F · P Q = −12 ft · lb. −→ 32. W = F · P Q= F −→ P Q cos 45◦ = (500)(100) √ √ 2/2 = 25,000 2 N · m 33. W = F ·15i = 15 · 50 cos 60◦ = 375 ft · lb. √ √ 34. W = F ·(15/ 3)(i + j + k) = −15/ 3 N · m 35. With the cube as shown in the diagram, and a the length of each edge, d1 = ai + aj + ak, d2 = ai + aj − ak, cos θ = (d1 · d2 ) / ( d1 d2 ) = 1/3, θ ≈ 71◦ z d2 y d1 x 36. Take i, j, and k along adjacent edges of the box, then 10i + 15j + 25k is along a diagonal, and a √ 2 3 5 unit vector in this direction is √ i + √ j + √ k. The direction cosines are cos α = 2/ 38, 38 38 38 √ √ ◦ cos β = 3/ 38, and cos γ = 5/ 38 so α ≈ 71 , β ≈ 61◦ , and γ ≈ 36◦ . Exercise Set 13.3 462 37. u + v and u − v are vectors along the diagonals, (u + v) · (u − v) = u · u − u · v + v · u − v · v = u 2 −v 2 so (u + v) · (u − v) = 0 if and only if u = v . 38. The diagonals have lengths u + v and u − v but u+v 2 = (u + v) · (u + v) = u 2 + 2u · v + v 2 , and u − v 2 = (u − v) · (u − v) = u 2 − 2u · v + v 2 . If the parallelogram is a rectangle then u · v = 0 so u + v 2 = u − v 2 ; the diagonals are equal. If the diagonals are equal, then 4u · v = 0, u · v = 0 so u is perpendicular to v and hence the parallelogram is a rectangle. u+v 2 = (u + v) · (u + v) = u 2 + 2u · v + v u−v 2 = (u − v) · (u − v) = u 2 − 2u · v + v , add to get u+v 39. 2 + u−v 2 =2 u 2 +2 v 2 and 2 2 The sum of th...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online