A block P weighing 200KN rests on the top of the block Q of weight 300KN and they are connected by a common string passing around a frictionless pulley S which is tied to a vertical wall by a horizontal string as shown if the coefficient of friction between the rubbing surfaces of the block P and block Q is 0.4 while that between the floor and the block Q is 0.3 determine the force F required to pull the block Q and the tension in each string

To solve this problem, we need to analyze the forces acting on the system.

1. First, let's consider block Q. The weight of block Q is 300 kN, which acts vertically downward. The normal force exerted by the floor on block Q is equal to its weight since there is no vertical acceleration. Therefore, the normal force exerted by the floor on block Q is also 300 kN.

The friction force acting on block Q is given by the equation: friction force = coefficient of friction * normal force.

So, the friction force between block Q and the floor is 0.3 * 300 kN = 90 kN.

2. Now, let's consider block P and the tension in the string. The weight of block P is 200 kN, which acts vertically downward. Since block P is in contact with block Q, there is a normal force between them. The normal force exerted by block Q on block P is equal to the weight of block P because there is no vertical acceleration.

The friction force between block P and block Q is given by the equation: friction force = coefficient of friction * normal force.

So, the friction force between block P and block Q is 0.4 * 200 kN = 80 kN.

Since the blocks are connected by a common string, the tension in the string is the same on both sides of the pulley.

3. To find the force F required to pull block Q, we need to consider the equilibrium of the system. The net force acting on block Q in the horizontal direction is zero because there is no horizontal acceleration.

The force F is equal to the tension in the string plus the friction force between block Q and the floor.

So, F = tension in the string + friction force between block Q and the floor.

Therefore, F = tension in the string + 90 kN.

Since the tension in the string is the same on both sides of the pulley, we can say that the tension in the string is equal to the friction force between block P and block Q:

Tension in the string = 80 kN.

Now, we can substitute this value into the equation for F:

F = 80 kN + 90 kN = 170 kN.

So, the force F required to pull block Q is 170 kN, and the tension in each string is 80 kN.