A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?
Given: The mass of the block is 25 kg , the mass of the man lifting the block is 50 kg and the normal force that the floor yields is 700 N .
First way to raise the mass is shown below:
Action on the floor by the man in the first case is,
F = ( m m + m b ) g
where, m m is the mass of the man and m b is the mass of the block.
By substituting the given values in the above equation, we get
F = ( ( 25 kg ) + ( 50 kg ) ) ( 10 m / s 2 ) = 750 N
Thus, the force that is applied on the floor by the man in the first case is 750 N .
The second way to raise the mass is shown below:
The equation of the action on the floor by the man in the second case is given as.
F = ( m m − m b ) g
By substituting the given values in the above equation, we get
F = ( ( 50 kg ) − ( 25 kg ) ) ( 10 m / s 2 ) = 250 N
Thus, the force that is applied on the floor by the man in the second case is 245 N .
As less force is applied on the floor by the man in the second case, the man should adopt the second method to raise the block.
Given: The mass of the block is
First way to raise the mass is shown below:
Action on the floor by the man in the first case is,
where,
By substituting the given values in the above equation, we get
Thus, the force that is applied on the floor by the man in the first case is
The second way to raise the mass is shown below:
The equation of the action on the floor by the man in the second case is given as.
By substituting the given values in the above equation, we get
Thus, the force that is applied on the floor by the man in the second case is
As less force is applied on the floor by the man in the second case, the man should adopt the second method to raise the block.
