Question

Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of 40 lb and the terminal point of the vector is point P(1, 1, 0). The second force has a magnitude of 60 lb and the terminal point of its vector is point Q(0, 1, 1). Let F be the resultant force of forces F1 and F2.
(a) Find the magnitude (in pounds) of F. (Round the answer to one decimal place.)
(b) Find the direction of F.

Answer 1

Answer:

Step-by-step explanation:

The two force F1 and F2  are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

$v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j$

The unit vector in the direction of force will be

$\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}$

The magnitude of the force is 40lb, so the force will be

$F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)$

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

$v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k$

$\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}$

The magnitude of the force is 60lb, so the force will be

$F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)$

The resultant of the two forces is

$F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k$

The magnitude force will be

$|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68$

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

$\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469$

The angle formed by F and y axis

$\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=$

The angle formed by F and z axis

$\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359$