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*4–108. Replace the force system by an equivalent resultant force and couple moment at point O. Take F3 = {-200i + 500j – 300k} N.
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*4–108. Replace the force system by an equivalent resultant force and couple moment at point O. Take F3 = {-200i + 500j – 300k} N.
Answer ( 1 )
Please briefly explain why you feel this answer should be reported.
We are given the force \( F_3 = (-200i + 500j – 300k) \, \mathrm{N} \).
We are required to replace the force system by an equivalent resultant force and couple moment acting at point \( O \).
The force \( F_1 \) is acting in the downward \( z \)-direction. The vector form of force \( F_1 \) can be written as,
\[
F_1 = (-300k) \, \mathrm{N}
\]
The force \( F_2 \) is acting in the positive \( y \)-direction. The vector form of force \( F_2 \) can be written as,
\[
F_2 = (200j) \, \mathrm{N}
\]
The equation for the equivalent resultant force acting at point \( O \) is given by,
\[
F_R = F_1 + F_2 + F_3
\]
Substitute all the values in the above equation.
\[
F_R = (-300k) \, \mathrm{N} + (200j) \, \mathrm{N} + (-200i + 500j – 300k) \, \mathrm{N}
\]
\[
F_R = (-200i + j(200 + 500) + k(-300 – 300)) \, \mathrm{N}
\]
\[
F_R = (-200i + 700j – 600k) \, \mathrm{N}
\]
The coordinate position of the point at which force \( F_1 \) is acting with respect to point \( O \) is given by
\[
P_1 = (0, 2, 0) \, \mathrm{m}
\]
The position vector from \( O \) to \( F_1 \) can be written as,
\[
V_1 = (0i + 2j + 0k) \, \mathrm{m}
\]
The coordinate position of the point at which force \( F_2 \) is acting with respect to point \( O \) is given by
\[
P_2 = (1.5, 3.5, 0) \, \mathrm{m}
\]
Substitute all the values in the couple moment equation.
\[
M_O =
\begin{vmatrix}
i & j & k \\
0 & 2 & 0 \\
0 & 0 & -300
\end{vmatrix}
+
\begin{vmatrix}
i & j & k \\
1.5 & 3.5 & 0 \\
0 & 200 & 0
\end{vmatrix}
+
\begin{vmatrix}
i & j & k \\
1.5 & 2 & 0 \\
-200 & 500 & -300
\end{vmatrix}
\]
Perform the determinant calculations:
\[
M_O = i\{- (300 \cdot 2) – (0 \cdot 0)\} – j\{(300 \cdot 0) – (0 \cdot 0)\} + k\{(0 \cdot 0) – (0 \cdot 2)\}
\]
\[
+ i\{(3.5 \cdot 0) – (0 \cdot 200)\} – j\{(1.5 \cdot 0) – (0 \cdot 0)\} + k\{(1.5 \cdot 200) – (3.5 \cdot 0)\}
\]
\[
+ i\{(-2 \cdot 300) – (500 \cdot 0)\} – j\{(1.5 \cdot 300) – (-200 \cdot 0)\} + k\{(1.5 \cdot 500) – (-200 \cdot 2)\}
\]
Simplify:
\[
M_O = (-600i) + (300k) + (-600i + 450j + 1150k) \, \mathrm{N \cdot m}
\]
\[
M_O = (-1200i + 450j + 1450k) \, \mathrm{N \cdot m}
\]