Answer:
The minimum mass of m2 before m1 starts sliding across the table is 57 grams
Explanation:
The given mass on the table, m1 = 250 g
The coefficient of friction between the table and the mass, m1, μ = 0.228
Therefore, the friction force, F between the mass m1 and the table is given as follows;
F = W₁ × μ
Where;
W₁ = The weight of m1 = Mass of m1 × Acceleration due to gravity
W₁ = 250 g × 9.81 m/s² = 0.25 kg × 9.81 m/s² = 2.4525 N
W₁ = 2.4525 N
Which gives;
F = 2.4525 N × 0.228 = 0.55917 N
F = 0.55917 N
By equilibrium of forces, the weight, W₂, of the mass m2 at which the mass m1 starts sliding is equal to the frictional force F of m1 or 0.55917 N
Therefore, for slipping to occur, we have;
W₂ = 0.55917 N
However, W₂ = Mass of m2 × Acceleration due to gravity
∴ W₂ = 0.55917 N = m2 × 9.81 m/s²
0.55917 N = m2 × 9.81 m/s²
m2 = 0.55917 N/(9.81 m/s²) = 0.057 kg = 57 grams
The minimum mass of m2 before m1 starts sliding across the table is 57 grams.
Answer:
3335400 N/m² or 483.75889 lb/in²
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
A = Area = 1.5 cm²
m = Mass of woman = 51 kg
F = Force = mg
When we divide force by area we get pressure
The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²