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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). - Physics

Sum

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.

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Solution


Thus, we have:
μR + μR = mg
⇒ 2μR = 40 × 10
`=>"R" = (40xx10)/(2xx0.8)=250 "N"`
Normal force = 250 N

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 6 Friction
Exercise | Q 30 | Page 99
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