A 60 kg man is standing on the 35 kg high table which is at rests on a horizontal floor. Determine the magnitude of the normal force that (i) the floor exerts on the table and (ii) the table exerts on the man.
(i) The normal force that the floor exerts on the table is equal in magnitude but opposite in direction to the weight of the table and the man combined. Therefore, we can calculate it as:
F_floor = m_table+man * g = (35 kg + 60 kg) * 9.81 m/s^2 = 931.35 N
So the floor exerts a normal force of 931.35 N on the table.
(ii) The normal force that the table exerts on the man is also equal in magnitude but opposite in direction to the weight of the man. Therefore, we can calculate it as:
F_table-man = m_man * g = 60 kg * 9.81 m/s^2 = 588.6 N
So the table exerts a normal force of 588.6 N on the man.