Which-of-the-following-set-of-horizontal-forces-would-lead-an-object-in-equilibrium-a-5N-10N-20N-b-6N-12N-18N-c-8N-8N-8N-b-2N-4N-8N-16N-

Question Number 107905 by aurpeyz last updated on 13/Aug/20

$$\mathrm{W}{hich}\:{of}\:{the}\:{following}\:{set}\:{of}\:{horizontal} \\ $$$${forces}\:{would}\:{lead}\:{an}\:{object}\:{in}\:{equilibrium}? \\ $$$$\left({a}\right)\:\mathrm{5}{N}\:\:\:\mathrm{10}{N}\:\:\mathrm{20}{N} \\ $$$$\left({b}\right)\:\mathrm{6}{N}\:\:\:\mathrm{12}{N}\:\:\:\:\mathrm{18}{N} \\ $$$$\left({c}\right)\:\mathrm{8}{N}\:\:\:\mathrm{8}{N}\:\:\:\mathrm{8}{N} \\ $$$$\left({b}\right)\:\mathrm{2}{N}\:\:\:\:\mathrm{4}{N}\:\:\:\mathrm{8}{N}\:\:\:\mathrm{16}{N} \\ $$

Commented by aurpeyz last updated on 13/Aug/20

$${pls}\:{help}\:{me}\:{with}\:{solution}\:{and}\:{explanation} \\ $$

Commented by mr W last updated on 13/Aug/20

Commented by mr W last updated on 13/Aug/20

only if all forces can form a closed shape. (b),(c) are possible

$${only}\:{if}\:{all}\:{forces}\:{can}\:{form}\:{a}\:{closed} \\ $$$${shape}. \\ $$$$\left({b}\right),\left({c}\right)\:{are}\:{possible} \\ $$

Answered by Don08q last updated on 13/Aug/20

The object will be in equilibrium if and only if one (the resultant) of the forces in any of the sets is equal to the algebraic sum of the others. In this case, that force must be directed to oppose the continuous directions of the other forces. (b) 6N + 12N = 18N (d) 2N + 4N + 8N = 16N

$$\mathrm{The}\:\mathrm{object}\:\mathrm{will}\:\mathrm{be}\:\mathrm{in}\:\mathrm{equilibrium}\:\mathrm{if} \\ $$$$\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{one}\:\left({the}\:{resultant}\right)\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{forces}\:\mathrm{in}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sets}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{algebraic}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{others}.\:\mathrm{In}\:\mathrm{this}\: \\ $$$$\mathrm{case},\:\mathrm{that}\:\mathrm{force}\:\mathrm{must}\:\mathrm{be}\:\mathrm{directed}\:\mathrm{to} \\ $$$$\mathrm{oppose}\:\mathrm{the}\:\mathrm{continuous}\:\mathrm{directions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{forces}. \\ $$$$\:\:\left({b}\right)\:\mathrm{6}{N}\:+\:\mathrm{12}{N}\:=\:\mathrm{18}{N} \\ $$$$\:\:\left({d}\right)\:\mathrm{2}{N}\:+\:\mathrm{4}{N}\:+\:\mathrm{8}{N}\:=\:\mathrm{16}{N}\: \\ $$$$ \\ $$

Commented by mr W last updated on 13/Aug/20

not correct sir! what you said is only valid if all forces are in the same direction. generally in 2D or in 3D, the condition is ΣF_i ^(→) =0, that means all forces should be able to form a closed polygon. besides 2+4+8=14≠16!

$${not}\:{correct}\:{sir}!\:{what}\:{you}\:{said}\:{is}\:{only} \\ $$$${valid}\:{if}\:{all}\:{forces}\:{are}\:{in}\:{the}\:{same} \\ $$$${direction}. \\ $$$${generally}\:{in}\:\mathrm{2}{D}\:{or}\:{in}\:\mathrm{3}{D}, \\ $$$${the}\:{condition}\:{is}\:\Sigma\overset{\rightarrow} {{F}_{{i}} }=\mathrm{0},\:{that}\:{means} \\ $$$${all}\:{forces}\:{should}\:{be}\:{able}\:{to}\:{form}\:{a} \\ $$$${closed}\:{polygon}. \\ $$$$ \\ $$$${besides}\:\mathrm{2}+\mathrm{4}+\mathrm{8}=\mathrm{14}\neq\mathrm{16}! \\ $$

Commented by mr W last updated on 13/Aug/20