Question

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 wall with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s2.
Figure

Answer 1

Answer:

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 wall with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s2.

Figure

Answer 2

(a) To Show: The person pushes the two wall with equal force.

(b) The normal force exerted by either wall on the person is 250 N

Explanation:

(a) The man pushes both walls, the walls in effect drive him in the same and opposite directions as well.  

Taking into account the horizontal forces on the man in horizontal direction.

Only the natural forces on the man will work on the walls. Let walls be N and N' for normal power. Then on top. N = N'

Therefore the man pushes both walls with equal forces and the forces of friction on both walls are equal to say F as well.  

(b) F+F = mg

2F = mg  

Where m is mass  

acceleration due to gravity, g

m = 40 kg  

 g = 10 $m/s^2$

2F = 40×10  

2F = 400  

 F = 400/2  

 F = 200  

Since the man is going to be in vertical balance,

μN = 200  

That the coefficient of friction between the body and the walls is 0.8 and that the limiting friction is  

applied at all touches.

So μ = 0.8

0.8N = 200  

N = 200/0.8  

N = 250