Question

A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm x 10 cm.
Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions
(a) 20 cm × 10 cm
⠀⠀⠀⠀and
(b) 40 cm × 20 cm.​

Answer 1

Given :-

• A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 10 cm.

To find :-

• The mass of the wooden block = 5 kg the dimensions

⠀⠀⠀= 40 cm × 20 cm × 10 cm

Here, the weight of the wooden block applies a thrust on the table top.

That is,

Solution :-

Thrust = F = m × g

⠀⠀⠀⠀⠀⠀⠀= 5 kg × 9.8 m s²

⠀⠀⠀⠀⠀⠀⠀= 49 N

Area of a side = Length × Breadth

⠀⠀⠀⠀⠀⠀⠀⠀⠀= 20 cm × 10 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀= 200 cm² = 0.02 m²

From Eq. (10.20),

$Pressure\ = \frac{49 \: N}{0.02 \: m²}$

⠀⠀⠀⠀⠀⠀⠀⠀= 2450 N m²

★ We know that ★

When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.

Area = Length × Breadth

⠀⠀⠀= 40 cm × 20 cm

⠀⠀⠀= 800 cm² = 0.08 m²

From Eq. (10.20),

$Pressure\ = \frac{49 \: N}{0.08 \: m²}$

⠀⠀⠀⠀⠀⠀⠀⠀= 612.5 N m²

Hence,

• The pressure exerted by the side 20 cm × 10 cm is 2450 N m² and by the side 40 cm × 20 cm is 612.5 N m².

Answer 2

Given:-

A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 10 cm.

To find:-

The mass of the wooden block = 5 kg the dimensions

⠀⠀⠀= 40 cm × 20 cm × 10 cm

Here, the weight of the wooden block applies a thrust on the table top.

That is,

Solution:-

Thrust = F = m × g

⠀⠀⠀⠀⠀⠀⠀= 5 kg × 9.8 m s²

⠀⠀⠀⠀⠀⠀⠀= 49 N

Area of a side = Length × Breadth

⠀⠀⠀⠀⠀⠀⠀⠀⠀= 20 cm × 10 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀= 200 cm² = 0.02 m²

From Eq. (10.20),

Pressure= $\sf\frac{49 \: N}{0.02 \: m²}$

⠀⠀⠀⠀⠀⠀⠀⠀= 2450 N m².