Question

Consider a force of 57.3 N, pulling 3 blocks of

Answer 1

Block 1 (the rightmost block) has

• net horizontal force

∑ F = F - T₁ - f₁ = m₁a

• net vertical force

∑ F = N₁ - m₁g = 0

where F = 57.3 N, T₁ is the tension in the string connecting blocks 1 and 2, f₁ is the magnitude of kinetic friction felt by block 1, m₁ = 0.8 kg is its mass, a is the acceleration you want to find, and N₁ is the magnitude of the normal force exerted by the surface.

Block 2 (middle) has much the same information:

• net horiz. force

∑ F = T₁ - T₂ - f₂ = m₂a

• net vert. force

∑ F = N₂ - m₂g = 0

with similarly defined symbols.

The same goes for block 3 (leftmost):

• net horiz. force

∑ F = T₂ - f₃ = m₃a

• net vert. force

∑ F = N₃ - m₃g = 0

We have m₁ = m₂ = m₃ = 0.8 kg, so I'll just replace each with m. It follows that each normal force has the same magnitude, N₁ = N₂ = N₃ = mg. And as a consequence of that, each frictional force has the same magnitude, f₁ = f₂ = f₃ = 0.4mg.

In short, the relevant equations are

[1] … 57.3 N - T₁ - 0.4mg = ma

[2] …T₁ - T₂ - 0.4mg = ma

[3] … T₂ - 0.4mg = ma

Adding [1], [2] and [3] together eliminates the tension forces, and we get

57.3 N - 1.2mg = 3ma

Solve for a :

57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) a

57.3 N - 9.408 N = (2.4 kg) a

a = (47.892 N) / (2.4 kg)

a ≈ 20.0 m/s²