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Question-228049




Question Number 228049 by Spillover last updated on 14/Mar/26
Answered by Kassista last updated on 14/Mar/26
in caseB, part of the force,Fsin (θ), is being used upward  which is useless, thefore, Person A is smarter
$${in}\:{caseB},\:{part}\:{of}\:{the}\:{force},{F}\mathrm{sin}\:\left(\theta\right),\:{is}\:{being}\:{used}\:{upward} \\ $$$${which}\:{is}\:{useless},\:{thefore},\:{Person}\:{A}\:{is}\:{smarter} \\ $$
Commented by fantastic2 last updated on 14/Mar/26
Fsin θ is not useless.it is reducing the  normal force ⇒reducing frictional force
$${F}\mathrm{sin}\:\theta\:{is}\:{not}\:{useless}.{it}\:{is}\:{reducing}\:{the} \\ $$$${normal}\:{force}\:\Rightarrow{reducing}\:{frictional}\:{force} \\ $$
Commented by TonyCWX last updated on 15/Mar/26
While in Case A, some of the forces is acting downwards.  So it′s wasted.
$$\mathrm{While}\:\mathrm{in}\:\mathrm{Case}\:\mathrm{A},\:\mathrm{some}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{is}\:\mathrm{acting}\:\mathrm{downwards}. \\ $$$$\mathrm{So}\:\mathrm{it}'\mathrm{s}\:\mathrm{wasted}. \\ $$
Commented by fantastic2 last updated on 15/Mar/26
the force is horizontal. no downward force
$${the}\:{force}\:{is}\:{horizontal}.\:{no}\:{downward}\:{force} \\ $$
Commented by TonyCWX last updated on 15/Mar/26
There will be some.    See how the fellow pushes that.
$$\mathrm{There}\:\mathrm{will}\:\mathrm{be}\:\mathrm{some}. \\ $$$$ \\ $$$$\mathrm{See}\:\mathrm{how}\:\mathrm{the}\:\mathrm{fellow}\:\mathrm{pushes}\:\mathrm{that}. \\ $$
Answered by fantastic2 last updated on 15/Mar/26
Commented by fantastic2 last updated on 15/Mar/26
f_A =μmg  f_B =μmg−μFsin α  F_A =F−μmg  F_B =Fcos α−μmg+μFsin α  F ?F(cos α+μsin α)
$${f}_{{A}} =\mu{mg} \\ $$$${f}_{{B}} =\mu{mg}−\mu{F}\mathrm{sin}\:\alpha \\ $$$${F}_{{A}} ={F}−\mu{mg} \\ $$$${F}_{{B}} ={F}\mathrm{cos}\:\alpha−\mu{mg}+\mu{F}\mathrm{sin}\:\alpha \\ $$$${F}\:?{F}\left(\mathrm{cos}\:\alpha+\mu\mathrm{sin}\:\alpha\right) \\ $$

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