Final answer:
The person pushes the walls with equal force due to Newton's Third Law of Motion and being in static equilibrium. The normal force exerted by either wall on the person can be calculated using the coefficient of friction and the person's weight.
Step-by-step explanation:
The question involves a person being at rest between two vertical walls by pressing them. This is a static equilibrium problem in physics where the forces exerted by the person on the walls must be equal and opposite due to Newton's Third Law of Motion. Since the person remains stationary, the net force is zero.
To show that the person pushes the two walls with equal force, we consider a physics principle: for the person to be in static equilibrium, the horizontal forces must balance each other. If one force were larger than the other, the person would move towards the weaker force, which is not the case.
To find the normal force exerted by either wall on the person, we can use the frictional force. The maximum frictional force before slipping occu rs is μN, where μ is the coefficient of friction and N is the normal force. As the person does not slip, we can equate the frictional force to the person's weight (W = mg) and solve for N. With a given coefficient of friction (μ = 0.8) and gravitational acceleration (g = 9.8 m/s²), and knowing that the person's weight (W) is the product of their mass (m = 40 kg) and g, we can calculate N. This is a classic statics problem that illustrates the principles of equilibrium and friction.