Hi I was facing some trouble solving this question. Can anyone explain the concept and help me solve this?

Mass of wooden box, m₁ = 8.60 kg

coefficient of kinetic friction with the inclined plane, μ₁ = = 0.35

Mass of cardboard box, m₂ = 1.30 kg

coefficient of kinetic friction between the two boxes , μ₂ = 0.45

angle of incline, θ = 38.0°

(a)

Let's consider the wooden box starts to slide downward along the incline with acceleration "a" 

Forces on wooden box are....

weight component downward along the incline, W₁ = m₁gsinθ

friction upward along the incline, f₁ = μ₁N₁ = μ₁(m₁gcosθ + m₂gcosθ) = (m₁+m₂)μ₁gcosθ

(Notice that the normal force (N₁) on wooden box will due to the weight of both the boxes

Forces on cardboard box are....

weight component downward along the incline, W₂ = m₂gsinθ

friction upward along the incline due to weight of m₂, f₂ = μ₂N₂ = μ₂m₂gcosθ

friction downward opposing the acceleration due to surface of wooden box, f₃ = μ₂N₂ = μ₂m₂gcosθ

(Notice that friction forces f₂ and f₃ have same magnitude but in opposite direction)

Let T₁ = tension on the rope attached to wooden box

and T₂ = tension on the rope attached to cardboard box

Net force, F = T₁ - T₂

(m₁+m₂)a = (W₁ - f₁) - (W₂ - f₂ + f₃)

(m₁+m₂)a = m₁gsinθ - (m₁+m₂)μ₁gcosθ - (m₂gsinθ - μ₂m₂gcosθ +  μ₂m₂gcosθ)

(m₁+m₂)a = m₁gsinθ - (m₁+m₂)μ₁gcosθ - m₂gsinθ 

(m₁+m₂)a = (m₁-m₂)gsinθ - (m₁+m₂)μ₁gcosθ 

a = g[(m₁-m₂)sinθ - (m₁+m₂)μ₁cosθ]/(m₁+m₂)

a = (9.81)[(8.60 - 1.30)sin38.0° - (8.60 + 1.30)(0.35)cos38.0°]/(8.60 + 1.30)

a = 1.75 m/s²

Since the rope connects both boxes, the wooden box has acceleration of 1.75 m/s² downward along the incline, while cardboard box has same acceleration of 1.75 m/s² but upward, along the incline, above the wooden box

(b)

If all surface are frictionless then 

Net force on the system, F = T₁ - T₂

F =  W₁ - W₂

F = m₁gsinθ - m₂gsinθ

F = (m₁ - m₂)gsinθ

F = (8.60 - 1.30)(9.81)sin38.0°

F = 44.1 N  downward along the incline

So we have to apply an opposite force of 44.1 N upward along the incline, to the wooden box to stop it from sliding 

Assume that the rope sections connecting to the boxes are parallel to the slope. If the pulley and rope are ideal, the tension (T) in the rope will be constant throughout its length.

Let's look at the top box first. Call it's mass m

The normal force of the lower box on the upper box will equal the weight component perpendicular to the slope and is 

N₁ = mgcosθ 

meaning the friction force between the boxes is 

F₁ = μ₁mgcosθ

If we assume that the lower box will tend to move down slope,

(This establishes our positive directions as up slope on the upper box and down slope on the lower box)

The upper box will tend to move up slope and the friction force will point down slope on the upper box and up slope on the lower box.

the F = ma equation acting parallel to the slope becomes

T - mgsinθ - μ₁mgcosθ = ma

T = ma + mgsinθ + μ₁mgcosθ

Now let's look at the bottom box

The normal force of the ramp on the bottom box is 

N₂ = (m + M)gcosθ

The friction force of the ramp on the bottom box is 

F₂ = μ₂(m + M)gcosθ

The F = ma equation becomes (down slope is positive direction)

                                                                       Mgsinθ - Τ - F₁ - F₂ = Ma

Mgsinθ - ma - mgsinθ - μ₁mgcosθ - μ₂(m + M)gcosθ - μ₁mgcosθ = Ma

                                      (M - m)gsinθ - gcosθ(2μ₁ + μ₂)m + μ₂M)  = (M + m)a

a = g((M - m)sinθ - cosθ((2μ₁ + μ₂)m + μ₂M))) / (M + m)

a = 9.81((8.60 - 1.30)sin38.0 - cos38.0((2(0.45) + 0.35)1.30 + 0.35(8.60))) / (8.60 + 1.30)

a = 0.111961... ≈ 0.112 m/s²◄

When the friction goes to zero, the equation becomes

(M + m)a = (M - m)gsinθ

If we want zero acceleration we add an unknown force T' assumed to be acting parallel to the slope and acting to prevent motion to the force side of the equation.

M + m)(0) = (M - m)gsinθ + T'

T' = (m - M)gsinθ

T' = (1.30 - 8.60)9.81sin38.0 = -44.0893... ≈ -44.1 N◄

The negative sign means the force is applied up slope on the bottom box or down slope on the top box.

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each box has a downward force of weight x sin 38 ± friction

 (friction force direction may vary depending on motions)

for the bottom box that is 8.6•9.8 sin 38 – 8.6•9.8•0.35•cos 38 = 51.89 – 23.24 = 28.65 N, downward direction

for the top box that is 1.3•9.8 sin 38 + 1.3•9.8•0.45•cos 38

(direction of friction force upwards because net motion is upwards)

    = 7.84 + 4.52 = 12.36

tension T:

for bottom box, F = ma = 28.65 – T

   8.6a = 28.65 – T

for top box, F = ma = T – 12.36

   1.3a = T – 12.36

adding

    9.9a = 28.65 – 12.36 = 16.29

   a = 1.64 m/s² (one upwards, one downwards)