1 Solution Set 1 on chapters 1&2&3

1 Solution Set 1 on chapters 1&2&3 - 1...

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N.B.: CJ th Edition. Chapter 1: Introduction and Mathematical Concepts Review the conversion of units, dimensional consistency, trigonometric functions, scalars and vectors, vector addition and subtraction, the components of a vector, addition of vectors by means of components. Solve problems CJ, pp. 22: # 6, 17, 44, 47, 62, 69 6. REASONING AND SOLUTION x has the dimensions of [L], v has the dimensions of [L]/[T], and a has the dimensions of [L]/[T] 2 . The equation under consideration is v n = 2 ax . The dimensions of the right hand side are L T L L T 2 2 2 = , while the dimensions of the left hand side are [ ] [ ] [ ] [ ] L L T T n n n = . The right side will equal the left side only when 2 n = . ________________________________________________________________________________ 17. REASONING AND SOLUTION Consider the following views of the cube. The length, L , of the diagonal of the bottom face of the cube can be found using the Pythagorean theorem to be L 2 = a 2 + a 2 = 2(0.281 nm) 2 = 0.158 nm 2 or L = 0.397 nm Side View Na Na Cl Bottom View Na Na Cl a a L L a c 1
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The required distance c is also found using the Pythagorean theorem. c 2 = L 2 + a 2 = (0.397 nm) 2 + (0.281 nm) 2 = 0.237 nm 2 Then, 0.487 nm c = 44. REASONING AND SOLUTION The force F can be first resolved into two components; the z component F z and the projection onto the x - y plane, F p as shown in the figure below on the left. According to that figure, F p = F sin 54.0° = (475 N) sin 54.0°= 384 N The projection onto the x-y plane, F p , can then be resolved into x and y components. 54.0° F z F z F p 54.0° F p F x F y y x 33.0° a. From the figure on the right, F x = F p cos 33.0° = (384 N) cos 33.0°= 322 N b. Also from the figure on the right, F y = F p sin 33.0° = (384 N) sin 33.0°= 209 N c. From the figure on the left, F z = F cos 54.0° = (475 N) cos 54.0°= 279 N ______________________________________________________________________________ 47. REASONING AND SOLUTION The horizontal component of the resultant vector is R h = A h + B h + C h = (0.00 m) + (15.0 m) + (18.0 m) cos 35.0° = 29.7 m Similarly, the vertical component is R v = A v + B v + C v = (5.00 m) + (0.00 m) + (– 18.0 m) sin 35.0° = – 5.32 m The magnitude of the resultant vector is 2
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R R R = + = + = h 2 v 2 m m 29 7 30 2 2 2 . –5.32 m . b g b g The angle θ is obtained from = tan –1 [(5.32 m)/(29.7 m)] = 10.2 ° ________________________________________________________________________________ 62. REASONING We will use the scalar x and y components of the resultant vector to obtain its magnitude and direction. To obtain the x component of the resultant we will add together the x components of each of the vectors. To obtain the y component of the resultant we will add together the y components of each of the vectors.
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1 Solution Set 1 on chapters 1&2&3 - 1...

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