Question

7. A mass m1 of 250 g is on a table connected to a massless pulley, as shown. The coefficient of friction between m1 and the table is 0.228. What is the maximum value of m2 before m1 starts sliding across the table?

Answer 1

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$_F$ between the mass m1 and the table is given as follows;

F$_F$ = 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$_F$ = 2.4525 N × 0.228 = 0.55917 N

F$_F$ = 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$_F$ 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.