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Question-203985




Question Number 203985 by ajfour last updated on 03/Feb/24
Commented by mr W last updated on 03/Feb/24
equilibrium is not possible!  ΣM_P ≠0
$${equilibrium}\:{is}\:{not}\:{possible}! \\ $$$$\Sigma{M}_{{P}} \neq\mathrm{0} \\ $$
Commented by mr W last updated on 03/Feb/24
Commented by ajfour last updated on 03/Feb/24
If we assume friction, can we   determine both tension and   existing friction, coefficient μ not  given but that equilibrium persists?
$${If}\:{we}\:{assume}\:{friction},\:{can}\:{we}\: \\ $$$${determine}\:{both}\:{tension}\:{and}\: \\ $$$${existing}\:{friction},\:{coefficient}\:\mu\:{not} \\ $$$${given}\:{but}\:{that}\:{equilibrium}\:{persists}? \\ $$
Commented by mr W last updated on 04/Feb/24
even with friction no equilibrium  is possible.  the resultant force (T_R  in diagram)  of the tension in string lies in the  direction of angle bisector. this force  and the weight (Mg) of the cylinder  always bring the cylinder to rotate  clockwise. the friction can not  prevent this and keep the cylinder   in equilibrium.  in the first case of following   diagrams an equilibrium is possible.  but we have the second case.
$${even}\:{with}\:{friction}\:{no}\:{equilibrium} \\ $$$${is}\:{possible}. \\ $$$${the}\:{resultant}\:{force}\:\left({T}_{{R}} \:{in}\:{diagram}\right) \\ $$$${of}\:{the}\:{tension}\:{in}\:{string}\:{lies}\:{in}\:{the} \\ $$$${direction}\:{of}\:{angle}\:{bisector}.\:{this}\:{force} \\ $$$${and}\:{the}\:{weight}\:\left({Mg}\right)\:{of}\:{the}\:{cylinder} \\ $$$${always}\:{bring}\:{the}\:{cylinder}\:{to}\:{rotate} \\ $$$${clockwise}.\:{the}\:{friction}\:{can}\:{not} \\ $$$${prevent}\:{this}\:{and}\:{keep}\:{the}\:{cylinder}\: \\ $$$${in}\:{equilibrium}. \\ $$$${in}\:{the}\:{first}\:{case}\:{of}\:{following}\: \\ $$$${diagrams}\:{an}\:{equilibrium}\:{is}\:{possible}. \\ $$$${but}\:{we}\:{have}\:{the}\:{second}\:{case}. \\ $$
Commented by mr W last updated on 04/Feb/24
Commented by mr W last updated on 04/Feb/24
Commented by ajfour last updated on 04/Feb/24
Yes true, unless θ=(π/2) equilibrium  not possible. We need have friction  at axle of upper cylinder and string.
$${Yes}\:{true},\:{unless}\:\theta=\frac{\pi}{\mathrm{2}}\:{equilibrium} \\ $$$${not}\:{possible}.\:{We}\:{need}\:{have}\:{friction} \\ $$$${at}\:{axle}\:{of}\:{upper}\:{cylinder}\:{and}\:{string}. \\ $$
Commented by mr W last updated on 04/Feb/24
yes sir!
$${yes}\:{sir}! \\ $$

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