Question 2.137: The force vector F points along the straight line from point......
The force vector F points along the straight line from point A to point B. Its magnitude is |F| = 20 N. The coordinates of points A and B are x_A = 6 m, y_A = 8 m, z_A = 4 m and x_B = 8 m, y_B = 1 m, z_B = – 2 m.
(a) Express the vector F in terms of its components.
(b) Use Eq. (2.34) to determine the cross products r_A × F and r_B × F.
Magnitude of force vector F: |F| = 20 N Coordinates of point A: x_A = 6 m, y_A = 8 m, z_A = 4 m Coordinates of point B: x_B = 8 m, y_B = 1 m, z_B = -2 m
Final Answer
We have r _A=(6 i +8 j +4 k )~m , r _B=(8 i + j -2 k )~m,
(b) \boxed{\begin{aligned}r _A \times F & =\frac{20~N }{\sqrt{89}}\left|\begin{array}{ccc} i & j & k \\6~m & 8~m & 4~m \\2 & -7 & -6\end{array}\right| \\& =(-42.4 i +93.3 j -123.0 k )~Nm \\r _B \times F & =\frac{20~N }{\sqrt{89}}\left|\begin{array}{ccc} i & j & k \\8~m & 1~m & -2~m \\2 & -7 & -6\end{array}\right| \\& =(-42.4 i +93.3 j -123.0 k )~Nm\end{aligned}}
Note that both cross products give the same result (as they must).