Question

The three vectors in Fig. 3-33 have magnitudes a = 3.00 m, b = 4.00 m, and c = 10.0 m and angle θ = 30.0°. What are

Answer 1

Answer:

a) $a_{x} =3$              b) $a_{y} =0$

c) $b_{x} =3.46$        d) $b_{y} =2$

e) $c_{x} =0$              f) $c_{y} =10$

g) p = -5.77                            h) q=5

Explanation:

Diagram for given question is attached below in fig 1

Part (a) (b)

for vector $\vec{a}$

θ = 0°

          $a_{x} = 3 cos (0)\\a_{x} = 3\\a_{y} = 3 sin (0)\\a_{y} = 0$

Part (c) (d)

for vector $\vec{b}$

θ = 30°

      $b_{x} = 4 cos (30)\\b_{x} = 3.46\\b_{y} = 4 sin (30)\\b_{y} = 2$

Part (e) (f)

for vector $\vec{c}$

θ = 90°

    $c_{x} = 10 cos (90)\\c_{x} = 0\\c_{y} = 10 sin (90)\\c_{y} = 10$

Part (g) (h)

                       $\vec{c} = p\vec{a} + q\vec{b}$

$c =c_{x} \hat{i} + c_{y}\hat{j}\\as a_{y} =0\\c_{x} \hat{i} + c_{y}\hat{j} = pa_{x} \hat{i} +q(b_{x} \hat{i} +b_{y} \hat{j} )\\c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\c_{y}\hat{j} =qb_{y}\hat{j}$

                  $q=\frac{c_{y}}{b_{y}} \\q=\frac{10}{2}\\q=5$

$c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\0 = p(3) + (5)(3.46)\\p = -5.77$