Question 2.163: The magnitude of the vertical force vector F in Problem 2.16......
The magnitude of the vertical force vector F in Problem 2.161 is 6 kN. Given that F + F_A + F_B + F_C = 0, what are the magnitudes of F_A,~F_B,~and~F_C?
The magnitude of the vertical force vector F in Problem 2.161 is 6 kN.
F + FA + FB + FC = 0.
rAD = 4i + 3j + 1k with a magnitude of |rAD| = √26 = 5.1
rBD = -1i + 3j - 2k with a magnitude of |rBD| = √14 = 3.74
rCD = -2i + 3j + 1k with a magnitude of |rCD| = √14 = 3.74
eAD = 0.7845i + 0.5883j + 0.1961k
eBD = -0.2673i + 0.8018j - 0.5345k
eCD = -0.5345i + 0.8018j + 0.2673k
FA = |FA|eAD
FB = |FB|eBD
FC = |FC|eCD
F = -6j (kN)
|FA| = 2.83 kN
|FB| = 2.49 kN
|FC| = 2.91 kN
So, the solution to the problem is that the magnitudes of the forces FA, FB, and FC are 2.83 kN, 2.49 kN, and 2.91 kN, respectively.
Final Answer
The strategy is to expand the forces into scalar components, and then use the force balance equation to solve for the unknowns. The unit vectors are used to expand the forces into scalar components. The position vectors, magnitudes, and unit vectors are:
\begin{aligned}& r _{A D}=4 i +3 j +1 k , \quad\left| r _{A D}\right|=\sqrt{26}=5.1 \\& e _{A D}=0.7845 i +0.5883 j +0.1961 k \\& r _{B D}=-1 i +3 j -2 k , \quad\left| r _{B D}\right|=\sqrt{14}=3.74 \\& e _{B D}=-0.2673 i +0.8018 j -0.5345 k \\& r _{C D}=-2 i +3 j +1 k , \quad\left| r _{C D}\right|=\sqrt{14}=3.74 \\& e _{C D}=-0.5345 i +0.8018 j +0.2673 k\end{aligned}The forces are:
F _A=\left| F _A\right| e _{A D}, F _B=\left| F _B\right| e _{B D}, F _C=\left| F _C\right| e _{C D}, F =-6 j~( kN ) .Substituting into the force balance equation
These simple simultaneous equations can be solved a standard method (e.g., Gauss elimination) or, conveniently, by using a commercial package, such as TK Solver®, Mathcad®, or other. An HP-28S hand held calculator was used here: |F_A| = 2.83 (kN), |F_B| = 2.49 (kN), |F_C| = 2.91 (kN)