B) I Ran Two Miles Due North Along the Lake. ______

B) I Ran Two Miles Due North Along the Lake. ______

<p>Name______Date______</p><p>DISCOVERY LAB: VECTORS </p><p>1. For the following questions, which of the following statements involve vector quantities, and which involve scalar quantities?</p><p> a) I ran two miles at the lake. ______</p><p> b) I ran two miles due north along the lake. ______</p><p> c) The plane flew at a speed of 90 knots on a heading of 600 east of north. ______</p><p> d) The car was traveling at 60 mph.______</p><p> e) The temperature was minus 20 degrees. ______</p><p> f) The cart moved 20 meters in the negative x direction. ______</p><p>2. Vector A points due west, while vector B points due south. They both have a magnitude of 4.0 cm. Using a ruler, </p><p>Draw the graphical addition of A + 2B to scale. Label A, 2B, (A + 2B).</p><p>Draw the graphical subtraction, (A – B) to scale. Label A, B, A - B.</p><p>Measure the length of each vector to 1 decimal place:</p><p>A + 2B ______A – B ______</p><p>1 3. Vectors A and B are shown in the drawing below. Vector R (not shown) is the resultant vector where R = A + B. Indicate the signs (negative or positive) of the following scalar components:</p><p>Ax ____Ay ____Bx ____ By ____ Rx ____ Ry, ____</p><p>Exercises 4-6: Draw and label x- and y- components of the vector shown:</p><p>Exercises 7-9: Draw and label x- and y- components of the vector and determine their numerical values (including sign!):</p><p>2 10. A common objective in physics is to calculate the components of the gravity vector parallel and perpendicular to a hill or incline. The vector F points in the “true vertical” direction (represented by the long end of the paper). The axes are tilted so that the x axis represents the hill’s orientation with respect to horizontal. a. Draw the F vector on the diagram of tilted coordinate axes below (F must remain true vertical at all times): b. Draw the components FX and FY on the diagram: c. Calculate the numerical value and sign of FX and FY:</p><p>FX = ______</p><p>FY = ______</p><p>Exercises 11-16: Draw and label the vector with these components. Write the magnitude of the vector below, along with the appropriate acute angle.</p><p>Ax = -3, Ay = 2 Bx = -3, By = -1 Cx = -2, Cy = 0 θ = ______θ = ______* θ = ______</p><p>* not acute in this case! </p><p>3 Ax = -1, Ay = 2 Bx = 0, By = -2 Cx = 3, Cy = -2</p><p>A = ______B = ______C = ______</p><p>θ = ______θ = ______θ = ______</p><p>Exercises 17-19: In each case, the vector A is 5 units long and is directed 0 30 above true horizontal. Determine the components Ax and Ay for the three coordinate systems shown below. Show your work below the figure.</p><p>Exercises 20-22 use graphical representation of vector addition. For each force, determine the x and y components by counting the number of squares to the nearest half. Include the appropriate “+” and “-“signs. Then, add the total number x and y squares from all the forces and draw the net force vector on the graph provided. </p><p>4 EXAMPLE x y</p><p>F1 (N) 0 2</p><p>F2 (N) -2.5 0</p><p>F3 (N) 0 -3</p><p>F4 (N) 3.5 -3.5</p><p>Fnet (N) 1 -4.5</p><p>#20 x y</p><p>F1 (N)</p><p>F2 (N)</p><p>F3 (N)</p><p>F4 (N)</p><p>Fnet (N)</p><p>#21 x y</p><p>F1 (N)</p><p>F2 (N)</p><p>F3 (N)</p><p>F4 (N)</p><p>Fnet (N)</p><p>#22 x y</p><p>F1 (N)</p><p>5 F2 (N)</p><p>F3 (N)</p><p>F4 (N)</p><p>Fnet (N)</p><p>Exercises 23-25 review the addition of vectors (in this case force vectors) using the mathematical procedure of vector addition. The procedure for all of these problems is shown below. </p><p>1. Enter the vector components in the table for each force vector. Put in the appropriate “+” and “ – ” signs based on quadrant. </p><p>2. Add components in the last row of the table.</p><p>3. Find the magnitude and acute angle above/below positive/negative x axis and enter it below the table.</p><p>Draw the net force vector on the graph provided.</p><p>EXAMPLE x component y component</p><p> o o F1 (N) 40 cos 45 40 sin 45 28.3 28.3</p><p>F2 (N) 0 30</p><p> o o F3 (N) -50 cos 30 -50 sin 30 -43.3 -25</p><p>Fnet (N) -15 33.3</p><p> o Fnet = 36.5 N Θ = 66 above –x axis</p><p>6 #23 x y component component</p><p>F1 (N)</p><p>F2 (N)</p><p>F3 (N)</p><p>Fnet (N)</p><p>Fnet = Θ =</p><p>#24</p><p>F1 (N)</p><p>F2 (N)</p><p>F3 (N)</p><p>Fnet (N)</p><p>Fnet = </p><p>#25 x component</p><p>F1 (N)</p><p>F2 (N)</p><p>F3 (N)</p><p>Fnet (N)</p><p>Fnet = Θ =</p><p>7</p>

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