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

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

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

Commented by aurpeyz last updated on 13/Aug/20

pls help me with solution and explanation

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

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

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

these three forces are in equilibrium.

$${these}\:{three}\:{forces}\:{are}\:{in}\:{equilibrium}. \\ $$

Commented by Don08q last updated on 13/Aug/20

Yes. Thanks for the correction

$${Yes}.\:{Thanks}\:{for}\:{the}\:{correction} \\ $$

Commented by aurpeyz last updated on 15/Aug/20

thanks for the answers

$${thanks}\:{for}\:{the}\:{answers} \\ $$

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