A brick of dimension 12m×10×5m is lying on the floor experience a force of 60N .Calculate the pressure.
Answers
We know, when force is constant, pressure ∞ 1/Area.
Therefore, pressure is maximum when area is minimum and vice versa.
By calculating the area of three faces, we get,
⇒A₁ = (20×10)= 200 sq m.
⇒A₂ = (10×5) = 50 sq m.
⇒A₃ = (20×5) = 100 sq m.
∴A₁ is maximum and A₂ is minimum.
∴Minimum pressure P₁ = F/A₁
= 30/200
= 0.15 Nm⁻².
∴Maximum pressure P₂ = F/A₂
= 30/50
= 0.6 Nm⁻². (Ans)
____________________♡
We know, when force is constant, pressure ∞ 1/Area.
Therefore, pressure is maximum when area is minimum and vice versa.
By calculating the area of three faces, we get,
⇒A₁ = (20×10)= 200 sq m.
⇒A₂ = (10×5) = 50 sq m.
⇒A₃ = (20×5) = 100 sq m.
∴A₁ is maximum and A₂ is minimum.
∴Minimum pressure P₁ = F/A₁
= 30/200
= 0.15 Nm⁻².
∴Maximum pressure P₂ = F/A₂
= 30/50
= 0.6 Nm⁻². (Ans)