A mass M is suspended by a rope, which goes around a pulley and is connected to mass m₁ = 3 kg which sits on an inclined plane. A mass m₂ = 2 kg sits on top of m₁ , as shown. The incline angle is = 17° . The pulley is frictionless, the rope is massless, and there is no friction between m₁ and the incline. However, there is friction between m₁ and m₂ , with coefficients µ_s = 0.4 and µ_k = 0.2 .

a) What is M if this system is in equilibrium (i.e. nothing is moving)?

b) What is the maximum value of M such that m₂ rides on top of m₁ without slipping?

a) Draw force diagrams; here, since m₁ and m₂ are not moving, we can treat them as one object of mass = m₁ + m₂. The rope is massless, so the tension from the diagram for mass M is the same as the tension for the mass on the incline. We have, from the hanging mass:

T - Mg = 0

(since we have a=0) or

T = Mg

and for the masses on the incline we have:

x:     T - (m₁ + m₂)gsin = 0

y:    N - (m₁ + m₂)gcos = 0

(again, since a=0). We don't need the y-equation right now; just plug in   T = Mg   from above to the x-equation to give

Mg = (m₁ + m₂)gsin

M = (m₁ + m₂)sin = (5 kg)sin(17°) = 1.46 kg       easy!

b) This is a bit harder; we now have to treat m₂ and m₁ separately. Draw force diagrams for each, and note that the frictional force on m₁ points down the incline, as it opposes motion, which means by the 3rd law, the frictonal force on m₂ must point up the incline (this is good, since we want to have m₂ go up the incline, and so there better be a force pointing that way!). Thus, for m₂

x:     F_f -m₂gsin = m₂a

y:     N₂ - m₂gcos = 0

        N₂ = m₂gcos

We have the usual inequality for the frictional force; since we want the maximum mass M, then we will need the maximum frictional force, so we can replace the inequality by the equality

F_f = µ_sN₂

where we use the static frictional coefficient, since the two surfaces will be at rest with respect to each other. Thus we can solve for the maximum acceleration that this maximum frictional force can provide:

F_f = µ_sN₂ = µ_s m₂gcos

hence

µ_s m₂gcos - m₂gsin = m₂a

a = µ_sgcos - gsin

For M we have

    T - Mg = Ma

or

T = M(g+a)

Finally, for the two-mass system, we have

T - (m₁ + m₂)gsin = (m₁ + m₂)a

similar to part a), except here   a   is non-zero. Plug in T = M(g+a), rearrange, and plug in   a = µ_sgcos - gsin    to get

M(g+a) = (m₁ + m₂)(a + gsin)

M(g+µ_sgcos - gsin ) = (m₁ + m₂)(µ_sgcos - gsin + gsin)

M = (m₁ + m₂)(µ_scos )/(1+µ_scos+µ_scos)

plug in the values for the masses, , and µ_s to get

M = 2.10 kg

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