fig 3.35 shows a brick of weight 2kgf and dimensions 20cm *10cm*5cm placed in three different positions on the ground.find the pressure exerted by the brick in each case
Answers
Answer:
To find the pressure exerted by the brick in each case, we need to use the formula:
Pressure = Force / Area
where the force is the weight of the brick and the area is the contact area between the brick and the ground.
First, let's calculate the weight of the brick in Newtons:
Weight = 2kgf * 9.81 m/s^2 = 19.62 N
Now, let's calculate the contact area between the brick and the ground for each case:
Case A:
The brick is placed on its largest surface, which has dimensions of 20cm x 10cm = 200cm^2 = 0.02 m^2.
Pressure = 19.62 N / 0.02 m^2 = 981 Pa
Therefore, the pressure exerted by the brick in Case A is 981 Pa.
Case B:
The brick is placed on one of its smaller surfaces, which has dimensions of 10cm x 5cm = 50cm^2 = 0.005 m^2.
Pressure = 19.62 N / 0.005 m^2 = 3924.8 Pa
Therefore, the pressure exerted by the brick in Case B is 3924.8 Pa.
Case C:
The brick is placed on one of its edges, which has dimensions of 5cm x 20cm = 100cm^2 = 0.01 m^2.
Pressure = 19.62 N / 0.01 m^2 = 1962 Pa
Therefore, the pressure exerted by the brick in Case C is 1962 Pa.