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प्रश्न
A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (in the following figure). Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two wall with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s2.
उत्तर
Thus, we have:
μR + μR = mg
⇒ 2μR = 40 × 10
`=>"R" = (40xx10)/(2xx0.8)=250 "N"`
Normal force = 250 N
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संबंधित प्रश्न
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The contact force exerted by a body A on another body B is equal to the normal force between the bodies We conclude that
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(b) the force of friction between the bodies is zero
(c) the magnitude of normal force equal that of friction
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