1 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 y right. 14 m x . (a) Use the inverse tangent function to find the tan 1. 34 or 34 below m distance angle : 14 m . r the +x axis 14 m m . 2 2. (b) Use the Pythagorean Theorem to r rx2 ry2 .. determine the magnitude of r : r 17 m 2 . (c) If both rx and ry are doubled, the direction tan 1 34 . 14 m 2 . will remain the same but the magnitude will 28 m 19 m 34 m 2 2. double: r .. 15. Picture the problem : The two vectors A (length 50 units) and B (length 120 units) are drawn y at right.. A. x Solution: 1. (a) Find Bx: Bx 120 units cos 70 41 units 70 .. 2. Since the vector A points entirely in the x direction, we can see that Ax = 50 units and that.
2 vector A has the greater x component.. B. 3. (b) Find By: Bx 120 units sin 70 113 units . 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 m) and B (length m) are drawn at right. y . B A. (a) A sketch (not to scale) of the vectors and their sum is shown at right.. C. x . (b) Add the x components: C x Ax Bx m cos m cos m Add the y components: C y Ay B y m sin m sin m . m m m 2 2. Find the magnitude of C : C Cx C y . 2 2. Cy 1 m . Find the direction of C : C tan 1 tan . Cx m . 3 1. 24. The vectors involved in the problem are depicted at right. y . Set the length of A B equal 37 A2 B 2 30.
3 2 2 2 . to 37 units: 37 A B A B.. B 37 2 A2 372 22 30 units B. 2. Solve for B: x 22 A O.. 29. The vector A has a length of 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) Multiply each component of A by : A m x .. A m x 23 m x 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. y m . (a) Find the direction of A from its A tan 1 22 . components: m A B . B. m m m 2 2. Find the magnitude of A : A A B . m m x O.. m A.. m . (b) Find the direction of B from its B tan 1 68 180 110 . components: m .. m m m 2 2. Find the magnitude of B : B .. (c) Find the components of A B : A B m x m y m x m y.
4 M . Find the direction of A B from its A B tan 1 45 . components: m .. Find the magnitude of A B : A B m 2 m 2 m 3 2.