A block of mass 10kg and dimensions 100cm x 20 cm x 50 cm rests on the floor with a force of 40N. Then the minimum pressure the block can exert on the floor is
Answers
Answer:
Given
Mass of wooden block, m = 10 kg
Dimensions of wooden block are 50 cm × 20 cm × 5 cm
To find
Ratio of the pressure exerted by the wooden block on the tabletop if it is made to lie on the table with its sides of dimension 20 cm × 5 cm and 50 cm × 20 cm
Formulae required
Newton's second Law
F = m a
Formula to calculate pressure
P = F / A
[ Where F is force, a is acceleration, m is mass, P is pressure, As area ]
Solution
Calculating the force applied by wooden block on the table ( F =? )
Using Formula for newton's second Law
→ F = m a
[ taking Mass of Block, m =10 kg and acceleration due to gravity, g = 9.8 m/s² ]
→ F = 10 × 9.8
→ F = 98 N
Evaluating pressure exerted by block when it is placed with sides 20 cm × 5 cm ( P₁ =? )
sides of dimensions with which it is placed on tabletop is 20 cm × 5 cm that is 0.2 m × 0.05 m
Using formula for calculating pressure
→ P₁ = F / A₁
→ P₁ = 98 / (0.2 × 0.05) Pascal
Evaluating pressure exerted by block when it is placed with sides 50 cm × 20 cm ( P₂ =? )
Sides of dimensions with which it is placed on tabletop is 50 cm × 20 cm that is 0.5 m × 0.2 m
Using formula for calculating pressure
→ P₂ = F / A₂
→ P₂ = 98 / (0.2 × 0.5) Pascal
Calculating the ratioof P₁ and P₂ ( P₁ : P₂ =? )
→ P₁ / P₂ = [98/(0.2×0.05)]/[98/(0.2×0.5)]
→ P₁ / P₂ = 0.5 / 0.05
→ P₁ / P₂ = 10 / 1
Therefore,
Ratio of the pressures exerted is 10 : 1.
Answer
Let P
1
,P
2
and P
3
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 (P
1
)
=
Area
Thrust
=
20×10
500
∴P
1
=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
(P
2
)=
Area
Thrust
=
20×5
500
∴P
2
=5gwtcm
−2
Case (iii) : When the block on 10cm×5cmface
Thrust= Weight of the brick =500g.wt.
Area of contact =10cm×5cm
Pressure=
Area
Thrust
=
10×5
500
P
3
=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.