Question 2.166: In Problem 2.165, let rBC be the position vector from point ......
In Problem 2.165, let r_{BC} be the position vector from point B to point C. Determine the cross product r_{BC} × T.
Position vector from point B to point C: r_BC
Cross product: r_BC × T
Step 1:
Write down the two vectors in the form of their components. In this case, we have:
Vector T = (333i + 333j - 167k) N Vector r_BC = (-0.35i + 0.2j - 0.2k) m
Vector T = (333i + 333j - 167k) N Vector r_BC = (-0.35i + 0.2j - 0.2k) m
Step 2:
Write down the determinant using the components of the vectors. The determinant is formed by arranging the unit vectors (i, j, k) in the first row, the components of the first vector (r_BC) in the second row, and the components of the second vector (T) in the third row.
Step 3:
Calculate the determinant. Multiply the components of the diagonals going from top left to bottom right and subtract the product of the diagonals going from top right to bottom left.
Step 4:
Simplify the determinant to obtain the cross product. The cross product will have components in the form of (i, j, k). In this case, the cross product is (33.3i - 125j - 183k) Nm.
Therefore, the cross product of vector r_BC and vector T is (33.3i - 125j - 183k) Nm.
Therefore, the cross product of vector r_BC and vector T is (33.3i - 125j - 183k) Nm.
Final Answer
From Problem 2.165 we know that
T = (333i + 333j – 167k) N
The vector r_{BC} is
r_{BC} = (- 035i + 0.2j – 0.2k) m
The cross product is
r _{B C} \times T =\left|\begin{array}{ccc} i & j & k \\-0.35 & 0.2 & -0.2 \\333 & 333 & -137\end{array}\right|=(33.3 i -125 j -183 k )~Nm
r _{B C} \times T =(33.3 i -125 j -183 k )~Nm
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