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Engineering Mechanics

D Alembert's Principle

A lift origin...

Question

A lift originally moving downwards at 10 m/s is brought to rest with a constant retardation in a distance of 25 m. The force with which the feet of a passenger of mass 80 kg (take g = 9.8 m/s $^{2}$ ) press downwards on the floor of the lift is

160 N

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784 N

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944 N

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1000 N

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Solution

The correct option is C 944 N
Applying D'Alembert's principle

$m (- a) = m g - R$

$∴ R = m g - m (- a)$
$= m (g + a) = 80 (9.81 + 2) = 944.8 N$

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A lift originally moving downwards at 10 m/s is brought to rest with a constant retardation in a distance of 25 m. The force with which the feet of a passenger of mass 80 kg (take g = 9.8 m/s2) press downwards on the floor of the lift is

The correct option is C 944 NApplying D'Alembert's principle m(−a)=mg−R ∴R=mg−m(−a) =m(g+a)=80(9.81+2)=944.8N

A lift originally moving downwards at 10 m/s is brought to rest with a constant retardation in a distance of 25 m. The force with which the feet of a passenger of mass 80 kg (take g = 9.8 m/s $^{2}$ ) press downwards on the floor of the lift is

The correct option is C 944 N
Applying D'Alembert's principle

$m (- a) = m g - R$

$∴ R = m g - m (- a)$
$= m (g + a) = 80 (9.81 + 2) = 944.8 N$