question_answer

A brick with dimensions of \[20\text{ }cm\text{ }\times \text{ }10\text{ }cm\text{ }\times \text{ }5\text{ }cm\]has a weight 500 g wt. Calculate the pressure exerted by it when it rests on different faces. In which case is the pressure exerted the greatest?

Answer:

Let\[{{P}_{1}},\text{ }{{P}_{2}}\]and \[{{P}_{3}}\] be the pressure exerted by the brick while resting on different faces. The dimensions of the given brick are 20 cm × \[10\text{ }cm\times 5\text{ }cm\] Case (i): When the block is resting on 20 cm × 10cm face. Thrust acting = Weight of the brick \[T=500\text{ }g\text{ }wt\] Area of contact \[(A)=20\text{ }cm\times 10\text{ }cm\] Pressure exerted \[({{P}_{1}})=?\] \[Pressure=\frac{Thrust}{Area}=\frac{500}{20\times 10}\] \[\Rightarrow \] \[{{P}_{1}}=2.5\text{ }g.wt\text{ }c{{m}^{2}}\] Case (ii): When placed on \[20\text{ }cm\times 5\text{ }cm\] face Thrust = Weight of the brick \[=500\text{ }g.\text{ }wt.\] Area of contact \[(A)=20\text{ }cm\times 5\text{ }cm\] Pressure exerted \[({{P}_{2}})=?\] \[Pressure=\frac{Thrust}{Area}=\frac{500}{20\times 5}\] \[\Rightarrow \] \[{{P}_{2}}=5\text{ }g.wt\text{ }c{{m}^{2}}\] Case (iii): When placed on\[10\text{ }cm\times 7\text{ }cm\]face 5 cm Thrust = Weight of the brick \[=500\text{ }g.\text{ }wt.\] Area of contact \[=10\text{ }cm\times 5\text{ }cm\] Pressure \[({{P}_{3}})=?\] \[Pressure=\frac{Thrust}{Area}=\frac{500}{10\times 5}\] \[\Rightarrow \] \[P=10\text{ }g.wt\text{ }c{{m}^{2}}\] 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 \[10\text{ }cm\times 5\text{ }cm\] face.