Question

Block 1, of mass m₁, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m₂, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is μ.

Part A

Find the ratio of the masses m₁ / m₂.

Answer 1

Part A Answer

To solve this problem, find the tension with respect to both masses, and then equate, since the tension is the same through the entire rope:

For mass 1:

(m1)g – T = (m1)a
T = (m1)g – (m1)a
T = (m1)(g – a)

For mass 2:

T – (m2)gsin(θ) – (m2)μgcos(θ) = (m2)a
T = (m2)a + (m2)gsin(θ) + (m2)μgcos(θ)
T = (m2)(a + gsin(θ) + μgcos(θ))

Now equate. Note how m1 and m2 were factored in the above equations. This was necessary to isolate m1/m2:

(m1)(g – a) = (m2)(a + gsin(θ) + μgcos(θ))
m1/m2 = (a + gsin(θ) + μgcos(θ)) / (g – a)

(a + gsin(θ) + μgcos(θ)) / (g – a)