Which will exert more pressure, $100 k g$ mass on $10 m^{2}$ or $50 k g$ mass on $4 m^{2}$ ? Show by actual calculation.

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Solution

$S t e p 1 : - G i v e n d a t a : - m_{1} = 100 k g A_{1} = 10 m^{2} m_{2} = 50 k g A_{2} = 4 m^{2} S t e p 2 : - T o f i n d : - P r e s s u r e (P) S t e p 3 : - W e k n o w n t h a t, g = 9.8 \frac{m}{s^{2}} F o r m u l a \Rightarrow P r e s s u r e = \frac{F o r c e}{A r e a} \Rightarrow F o r c e = m a s s \times a c c e l e r a t i o n d u e g r a v i t y (g) F = m \times g S t e p 4 : - C o n d i t i o n 1 : - \Rightarrow P_{1} = \frac{F_{1}}{A_{1}} \Rightarrow P_{1} = \frac{m_{1} \times g}{A_{1}} \Rightarrow P_{1} = \frac{100 \times 9.8}{10} \Rightarrow P_{1} = \frac{980}{10} \Rightarrow P_{1} = 98 \frac{N}{m^{2}} C o n d i t i o n 2 : - \Rightarrow P_{2} = \frac{F_{2}}{A_{2}} \Rightarrow P_{2} = \frac{m_{2} \times g}{A_{2}} \Rightarrow P_{2} = \frac{50 \times 9.8}{4} \Rightarrow P_{2} = \frac{490}{4} \Rightarrow P_{2} = 122.5 \frac{N}{m^{2}} S t e p 5 : - \Rightarrow P_{1} = 98 \frac{N}{m^{2}} & P_{2} = 122.5 \frac{N}{m^{2}} \Rightarrow H e n c e, P_{2} e x e r t m o r e p r e s s u r e t h a n P_{1}$

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