A 60 kg man is standing on the 35 kg high table which is at rest on a horizontal floor. Determine the magnitude of the normal force that
(i) the floor exerts on the table and (il) the table exerts on the man.
To calculate the normal force, we need to consider the forces acting on each object in the vertical direction.
Let's start with the normal force the floor exerts on the table.
(i) Normal force from the floor on the table:
Since the table is at rest on a horizontal floor, the net force acting on the table in the vertical direction must be zero.
Let N1 be the normal force exerted by the floor on the table.
Considering only the forces acting in the vertical direction:
N1 - Weight of the table = 0
N1 - (mass of table * gravity) = 0
N1 = 35 kg * 9.8 m/s^2
N1 = 343 N
Therefore, the normal force that the floor exerts on the table is 343 N.
Now, let's calculate the normal force that the table exerts on the man.
(ii) Normal force from the table on the man:
Similar to the previous calculation, the net force acting on the man must also be zero since he is at rest on the table.
Let N2 be the normal force exerted by the table on the man.
Considering only the forces acting in the vertical direction:
N2 - Weight of the man - Weight of the table = 0
N2 - (mass of man * gravity) - (mass of table * gravity) = 0
N2 = (mass of man + mass of table) * gravity
N2 = (60 kg + 35 kg) * 9.8 m/s^2
N2 = 941 N
Therefore, the normal force that the table exerts on the man is 941 N.