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three-forces-F-1-F-2-and-F-3-acts-through-the-points-with-position-vectors-r-1-r-2-and-r-3-respectively-where-F-1-3i-2j-4k-N-r-1-i-k-m-F-2-i-j-N-r-2-j-k-m-F-3-i-4k-N-r-3-




Question Number 65170 by Rio Michael last updated on 25/Jul/19
three forces F_1 , F_2  and F_3  acts through the points with position vectors  r_1 ,r_2  and r_3  respectively where   F_1  =(3i −2j−4k)N,   r_1 = (i +k)m  F_2 =(−i+j)N,  r_2 =(j+k)m  F_3 =(−i+4k)N, r_3 =(i+j+k)m  a. show that this system does not reduce to a single force.   When a fourth force F is added, the system of forces is in equilibrium  b. Show that F acts through  the point with vectors (3k)m.
$${three}\:{forces}\:{F}_{\mathrm{1}} ,\:{F}_{\mathrm{2}} \:{and}\:{F}_{\mathrm{3}} \:{acts}\:{through}\:{the}\:{points}\:{with}\:{position}\:{vectors} \\ $$$$\boldsymbol{{r}}_{\mathrm{1}} ,{r}_{\mathrm{2}} \:{and}\:{r}_{\mathrm{3}} \:{respectively}\:{where} \\ $$$$\:{F}_{\mathrm{1}} \:=\left(\mathrm{3}{i}\:−\mathrm{2}{j}−\mathrm{4}{k}\right){N},\:\:\:{r}_{\mathrm{1}} =\:\left({i}\:+{k}\right){m} \\ $$$${F}_{\mathrm{2}} =\left(−{i}+{j}\right){N},\:\:{r}_{\mathrm{2}} =\left({j}+{k}\right){m} \\ $$$${F}_{\mathrm{3}} =\left(−{i}+\mathrm{4}{k}\right){N},\:{r}_{\mathrm{3}} =\left({i}+{j}+{k}\right){m} \\ $$$${a}.\:{show}\:{that}\:{this}\:{system}\:{does}\:{not}\:{reduce}\:{to}\:{a}\:{single}\:{force}. \\ $$$$\:{When}\:{a}\:{fourth}\:{force}\:{F}\:{is}\:{added},\:{the}\:{system}\:{of}\:{forces}\:{is}\:{in}\:{equilibrium} \\ $$$${b}.\:{Show}\:{that}\:{F}\:{acts}\:{through}\:\:{the}\:{point}\:{with}\:{vectors}\:\left(\mathrm{3}{k}\right){m}. \\ $$

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