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 (figure 6-E11). 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 walls with equal force. (b) Find the normal force exerted by either wall on the person.
Take g = 10 m/s2.
Mass of man = 50kg
g = 10 m/s2
Frictional force developed between hands, legs & back with the wall will be equal to the weight of the man. So he remains in equilibrium.
He gives equal force on both the walls and gets equal reaction R from both the wall (Newton’s third law of motion). If he applies unequal forces on the wall, the reaction force will be different and he can’t rest between the walls. Frictional force 2μR balance his body weight. From the free body diagram
μR + μR = mg
⇒ 2μR = 40 × 10
⇒R=40×10/2×0.8=250 N
Normal force = 250 N