Physics Solution Manual for 1100 and 2101

# Physics Solution Manual for 1100 and 2101

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Unformatted text preview: n by the 2 2 2 Pythagorean theorem: F = A + B . Knowing the magnitudes of A and B, we can calculate the magnitude of F. The direction of the resultant can be obtained using trigonometry. b. For the vector F′ = A – B we note that the subtraction can be regarded as an addition in the following sense: F′ = A + (–B). The vector –B points due south, opposite the vector B, so the two vectors are once again perpendicular and the magnitude of F′ again is given by the Pythagorean theorem. The direction again can be obtained using trigonometry. Chapter 1 Problems SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have F 2 = A2 + B 2 F= F B θ 445 325 b N g+ b N g θ F′ A 2 North North A2 + B 2 F= 15 –B A 2 (b) (a) = 551 N Using trigonometry, we can see that the direction of the resultant is tan θ = B A or θ = tan −1 325 F NI= G NJ HK 445 36.1° north of west b. Referring to the drawing and following the same procedure as in part a, we find bg F ′2 = A2 + − B 2 tan θ = F′ = or B A or 445 − b g = b N g+ b325 N g = 325 F N I = 36.1° south of west G NJ HK 445 A2 + − B θ = tan −1 2 2 2 551 N _____________________________________________________________________________________________ 26. REASONING The triple jump consists of a double jump in one direction, followed by a perpendicular single jump, which we can represent with displacement vectors J and K (see the drawing). K These two perpendicular vectors form a right triangle with their resultant D = J + K, which is the displacement of the colored D checker. In order to find the magnitude D of the displacement, we first need to find the magnitudes J and K of the double jump and the single jump. As the three sides of a right triangle, J, K, and D (the hypotenuse) are related to one another by the Pythagorean ds J theorem (Equation 1.7) The double jump moves the colored s checker a straight-line distance equal to the length of four square’s diagonals d, and the single jump moves a length equal to two square’s diagonals. Therefore, J = 4d and K = 2d (1) Let the length of a square’s side be s. Any two adjacent sides of a square form a right triangle with the square’s diagonal (see the drawing). The Pythagorean theorem gives the diagonal length d in terms of the side length s: 16 INTRODUCTION AND MATHEMATICAL CONCEPTS d = s2 + s2 = 2s2 = s 2 (2) SOLUTION First, we apply the Pythagorean theorem to the right triangle formed by the three displacement vectors, using Equations (1) for J and K: D= ( 4d )2 + ( 2d )2 J 2 + K2 = = 16d 2 + 4d 2 = 20d 2 = d 20 (3) Substituting Equation (2) into Equation (3) gives ( D = d 20 = s 2 ) 20 = s 40 = ( 4.0 cm ) 40 = 25 cm 27. REASONING For convenience, we can assign due east to be the positive direction and due west to be the negative direction. Since all the vectors point along the same east-west line, the vectors can be added just like the usual algebraic addition of positive and negative scalars. We will carry out the addition for all of the possible choices for the two vectors and identify the resultants with the smallest and largest magnitudes. SOLUTION There are six possible choices for the two vectors, leading to the following resultant vectors: F1 + F2 = 50.0 newtons + 10.0 newtons = +60.0 newtons = 60.0 newtons, due east F1 + F3 = 50.0 newtons − 40.0 newtons = +10.0 newtons = 10.0 newtons, due east F1 + F4 = 50.0 newtons − 30.0 newtons = +20.0 newtons = 20.0 newtons, due east F2 + F3 = 10.0 newtons − 40.0 newtons = −30.0 newtons = 30.0 newtons, due west F2 + F4 = 10.0 newtons − 30.0 newtons = −20.0 newtons = 20.0 newtons, due west F3 + F4 = −40.0 newtons − 30.0 newtons = −70.0 newtons = 70.0 newtons, due west The resultant vector with the smallest magnitude is F1 + F3 = 10.0 newtons, due east . The resultant vector with the largest magnitude is F3 + F4 = 70.0 newtons, due west . Chapter 1 Problems 28. REASONING Both P and Q and the vector sums K and M can be drawn with correct magnitudes and directions by counting grid squares. To add vectors, place them tail-to-head and draw the resultant vector from the tail of the first vector to the head of the last. The vector 2P is equivalent to P + P, and −Q is a vector that has the same magnitude as Q, except it is directed in the opposite direction. 17 P Q M −Q 8.00 cm K P The vector M runs 11 squares P horizontally and 3 squares vertically, and the vector K runs 4 squares horizontally and 9 squares vertically. These distances can be converted from grid squares to centimeters with the grid scale: 1 square = 4.00 cm. Once the distances are calculated in centimeters, the Pythagorean theorem (Equation 1.7) will give the magnitudes of the vectors. SOLUTION a. The vector M = P + Q runs 11 squares horizontally and 3 squares vertically, and these cm distances are equivalent to, respectively, 4.00 (11 squares ) = 44.0 cm and square cm 4.00 ( 3 squares ) = 12.0 cm . Thus, the magnitude of M is...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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