Vectors in Physics

Chapter 3: Solutions of Homework Problems Vectors in Physics  12. Picture the Problem: The given vector components correspond to the vector r as drawn at right. (a) Use the inverse tangent function to find the distance angle  :  9.5m    tan    34 or 34° below  14 m  the +x axis (b) Use the Pythagorean Theorem to  determine the magnitude of r : r  rx2  ry2  (c) If both rx and ry are doubled, the direction will remain the same but the magnitude will double:   tan 1  y 14 m x 1 14 m  2   9.5 m  −9.5 m  r 2 r  17 m  9.5m  2    34  14 m  2  r  28 m  2   19 m   34 m 2   15. Picture the Problem: The two vectors A (length 50 units) and B (length 120 units) are drawn at right. y  A Bx  120 units  cos 70  41 units Solution: 1. (a) Find Bx:  2. Since the vector A points entirely in the x direction, we can see that Ax = 50 units and that  vector A has the greater x component.  B Bx  120 units  sin 70  113 units 3. (b) Find By: x 70°   4. The vector A has no y component, so it is clear that vector B has the greater y component. However, if one takes  into account that the y-component of B is negative, then it follows that it smaller than zero, and hence A has the greater y-component.   20. The two vectors A (length 40.0 m) and B (length 75.0 m) are drawn at right. y  B (a) A sketch (not to scale) of the vectors and their sum is shown at right.  A 20.0° (b) Add the x components: C x  Ax  Bx   40.0 m  cos  20.0    75.0 m  cos  50.0    85.8 m Add the y components: C y  Ay  B y   40.0 m  sin  20.0    75.0 m  sin  50.0    43.8 m  2 2 Find the magnitude of C : C  Cx  C y   Find the direction of C :  85.8 m  2   43.8 m   96.3 m 2  Cy  1  43.8 m    tan    27.0  85.8 m   Cx   C  tan 1  3–1  C 50.0° x 24. The vectors involved in the problem are depicted at right. y   Set the length of A  B equal to 37 units: 37  A  B Solve for B: B  37 2  A2  372   22   30 units 37  2 A2  B 2 2 30   AB 2  B 2  A −22 29. O x  The vector A has a length of 6.1 m and points in the negative x direction. Note that in order to multiply a vector by a scalar, you need only multiply each component of the vector by the same scalar.   A    6.1 m  xˆ (a) Multiply each component of A by −3.7:  3.7 A    3.7   6.1 m   xˆ   23 m  xˆ so Ax  23 m  (b) Since A has only one component, its magnitude is simply 23 m. 31. Picture the Problem: The vectors involved in the problem are depicted at right.  (a) Find the direction of A from its components:  A  tan 1   Find the magnitude of A : A y  2.0 m    –22  5.0 m   5.0 m  2   AB   2.0 m   5.4 m 2 2.0 m O 2.0 m  A  B  A  (b) Find the direction of B from its components:  B  tan 1   Find the magnitude of B : B   (c) Find the components of A  B :   A  B   5.0  2.0 m  xˆ   2.0  5.0 m  yˆ   3.0 m  xˆ   3.0 m  yˆ   Find the direction of A  B from its components:  A  B  tan 1    Find the magnitude of A  B :   AB   5.0 m    68  180  110  –2.0 m   2.0 m  2   5.0 m   5.4 m 2  3.0 m    45  3.0 m   3.0 m 2   3.0 m 2 3–2  4.2 m  B 5.0 m x