The free body diagram of the mass 2m is shown in Figure 1.

The free body diagram of the mass m is shown in Figure 2.

The free body diagram of the mass M is shown in Figure 3.

Apply Newton’s second law in the Figure1.

    T1=f1+2m(gsinθ+a)        (I)

Here, T1 is the tension acting on the mass 2m, m is the mass, g is the acceleration due to gravity, f1  frictional force on the mass 2m, a is the acceleration, and θ is the angle of inclination.

Apply Newton’s second law in the Figure 2.

    T2−T1=f2+m(gsinθ+a)        (II)

Here, T2 is the tension acting on the mass m, and f2 is the frictional force on mass m.

Apply Newton’s second law in the Figure 3.

    T2=M(g−a)        (III)

Assume that, the system is in equilibrium, a=0, and the incline is friction less f1=f2=0.

Use equation (I) and (III), in (II).

    Mg−2mgsinθ=mgsinθM=3msinθ        (IV)

Conclusion:

Therefore, the mass M in terms of m, g, and θ is 3msinθ_.

From the part (a), T1 is 2mgsinθ when the system is in equilibrium.

Use equation (IV) in (III), and substitute 0 for a and f1.

    T2=3mgsinθ

Conclusion:

Therefore, the tensions T1 is 2mgsinθ_, and T2 is 3mgsinθ_.

Assume that, M=6msinθ, and f1=f2=0.

Substitute, 6msinθ for M in the equation (III).

    T2=6msinθ(g−a)        (V)

Use equation (I) and (V) in (VII), substitute, 0 for f1, and f2, and solve for a.

    6msinθ(g−a)−2m(gsinθ+a)=m(gsinθ+a)6mgsinθ−6msinθa−2mgsinθ−2ma=mgsinθ+ma3gsinθ−3a−6sinθa=0a=3gsinθ3+6sinθ=gsinθ1+2sinθ        (VI)

Conclusion:

Therefore, the acceleration of each object when the value of M is doubled is a=gsinθ1+2sinθ_.

Use equation (VI) in (I), substitute, 0 for f1.

    T1=0+2m(gsinθ+gsinθ1+2sinθ)=2mgsinθ(1+11+2sinθ)=2mgsinθ(1+2sinθ+11+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)        (VII)

Use equation (VI) in (III), and use 6msinθ for M.

    T2=6msinθ(g−gsinθ1+2sinθ)=6mgsinθ(1−sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)        (VIII)

Conclusion:

Therefore, the tensions T1 is T1=4mgsinθ(1+sinθ1+2sinθ)_ and T2 is T2=6mgsinθ(1+sinθ1+2sinθ)_ when the value of M is doubled.

At the equilibrium the acceleration is zero, and the frictional forces acting down the incline.

Use 2μsmgcosθ for f1, and 0 for a in the equation (I).

    T1=2μsmgcosθ+2mgsinθ=2mg(sinθ+μscosθ)        (IX)

Use μsmgcosθ for f2, and 0 for a in the equation (II).

    T2−T1=μsmgcosθ+mgsinθ=mg(sinθ+μscosθ)        (X)

Rewrite the equation (III) for Mmax.

    T2,max=Mmaxg        (XI)

Use equation (IX) and (X) in (XI).

    Mmaxg−2mg(sinθ+μscosθ)=mg(sinθ+μscosθ)Mmaxg=2mgsinθ+2mgμscosθ+mgsinθ+mgμscosθMmax=3msinθ+3mμscosθ=3m(sinθ+μscosθ)

Conclusion:

Therefore, the maximum value of M at the equilibrium is Mmax=3m(sinθ+μscosθ)_.

Rewrite the equation (III) for Mmin.

    T2,min=Mming        (XII)

At the equilibrium the acceleration is zero, and impending motion up the incline, so the frictional forces acting up the incline.

Use 2μsmgcosθ for f1, and 0 for a in the equation (I).

    T1=2μsmgcosθ−2mgsinθ=2mg(sinθ−μscosθ)        (XIII)

Since the frictional forces acting up the incline.

Use μsmgcosθ for f2, and 0 for a in the equation (II).

    T2−T1=μsmgcosθ−mgsinθ=mg(sinθ−μscosθ)        (XIV)

Since the frictional forces acting up the incline.

Use equation (XII) and (XIII) in (XIV).

    Mming−2mg(sinθ−μscosθ)=mg(sinθ−μscosθ)Mming=2mgsinθ−2mgμscosθ+mgsinθ−mgμscosθMmin=3msinθ−3mμscosθ=3m(sinθ−μscosθ)        (XV)

Conclusion:

Therefore, the minimum value of M at the equilibrium is Mmin=3m(sinθ−μscosθ)_.

Subtract equation (XII) from (XI).

    T2,max−T2,min=Mmaxg−Mming        (XVI)

Use 3m(sinθ+μscosθ) for Mmax, and 3m(sinθ−μscosθ) for Mmin in the equation (XVI).

    T2,max−T2,min=3mg(sinθ+μscosθ)−3mg(sinθ−μscosθ)=3mgμscosθ+3mgμscosθ=6mgμscosθ

Conclusion:

Therefore, the value of T2 when the M has its maximum and minimum is 6μsmgcosθ_.