a brick weighs 4 kgf having dimensions of 20 cm ×10cm× 5cm . how should you place the brick so that it has least pressure. what is the pressure exerted
Answers
Answer:
Let P1,P2 and P3 be the pressures exerted by the brick while resting on different faces.
The dimensions of the given brick are 20cm×10cm×5cm
Case (i) : When the block is resting on 20cm×10cmface.
Thrust acting= Weight of the brick
T=500gwt
Area of constant (A)=20cm×10cm
Pressure exerted (P1)
=AreaThrust=20×10500
∴P1=2.5gwtcm−2
Case (ii) : When the block is resting on 20cm×5cmface
Thrust= Weight of the brick
=500gwt
Area of constant (A)=20cm×5cm
Pressure exerted
(P2)=AreaThrust=20×5500
∴P2=5gwtcm−2
Case (iii) : When the block on 10cm×5cmface
Thrust= Weight of the brick =500g.wt.
Area of contact =10cm×5cm
Pressure=AreaThrust=10×5500
P3=10gwtcm−2
∴ From the above three cases, it is clear that, as the area of contact decreases, the pressure exerted increases and is greater when the brick rests on its 10cm×5cm face.
Answer:
5 cm as height
2kPa (1.96 kPa)
Explanation:
Answer 1:-
The brick should be placed in such a way that the edge with 5 cm is the height. As doing this the side of brick with maximum surface area will be exposed to ground and the pressure will be minimum.
Answer 2:-
Area of the side with dimensions 20 cm and 10 cm = 20 * 10
= 200cm²
= 200/10000 m²
= 0.02 m²
Force in newton (g = 10m/s²) = 4 kgf * 10
= 40 N
Pressure due to brick = 40N/0.02m²
= 2000N/m² or 2kPa
Force in newton (g = 9.8m/s²) = 4kgf * 9.8
= 39.2 N
Pressure due to brick = 39.2N/0.02m²
= 1960N/m² or 1.96 kPa