Verified answer

The following assumptions are made:

1. Motion of the system is downward along the incline

2. The pulley (represented by the mustard-colored circle) is ideal, hence frictionless.

For the BLUE block,

Summation of forces along the x - axis (parallel to the incline)

Mb(g)sin (37) - T1 - Fb = Mb(a) -- call this Equation 1

Summation of forces along the y - axis (perpendicular to the incline)

Mb(g)(cos 37) = Nb -- call this Equation 2

where

Mb = mass of the blue block = 2 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

T1 = tension on the string between the blue and red blocks

Fb = friction force acting on the blue block

a = acceleration of the system

Nb = normal force acting on the blue block

By definition,

Fb = (mu)(Nb) and therefore, Equation 2 becomes

Fb = (mu)(Mb)(g)(cos 37) and substituting this in Equation 1,

Mb(g)sin (37) - T1 - (mu)(Mb)(g)(cos 37) = Mb(a)

Mb(g)[sin 37 - (mu)(cos 37)] - T1 = Mb(a)

Solving for T1,

T1 = Mb(g)[sin 37 - (mu)(cos 37)] - Mb(a) -- call this Equation 4

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For the RED block

Summation of forces along the x - axis (parallel to the incline)

T1 + Mr(g)sin 37 - T2 - Fr = Mr(a) --- call this Equation 5

Summation of forces along the y - axis (perpendicular to the incline)

Mr(g)(cos 37) = Nr -- call this Equation 6

where

Mr = mass of the red block = 5 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

T1 = tension on the string between the blue and red blocks

Fr = friction force acting on the red block

a = acceleration of the system

Nr = normal force acting on the blue block

Again, by definition,

Fr = (mu)(Nr) and therefore, Equation 6 becomes

Fr = (mu)(Mr)(cos 37) and substituting this in Equation 5,

T1 + Mr(g)sin 37 - T2 - (mu)(Mr)(cos 37) = Mr(a)

From Equation 4, substituting T1 in the above,

Mb(g)[sin 37 - (mu)(cos 37)] - Mb(a) + Mr(g)sin 37 - T2 - (mu)(Mr)(cos 37) = Mr(a)

and solving for "T2"

T2 = Mb(g)[sin 37 - (mu)(cos 37)] - Mb(a) + Mr(g)sin 37 - (mu)(Mr)(cos 37) - Mr(a)

Simplifying the above,

T2 = Mb(g)[sin 37 - (mu)(cos 37)] + Mr(g)[sin 37 - (mu)(cos 37)] - a(Mb + Mr) --- call this Equation 7

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For the GREEN block

T3 - Mg(g) = Mg(a)

T3 = Mg(a) + Mg(g)

where

T3 = tension in the string that is pulling the green block

Mg = mass of the green block = 1 kg

g = 9.8 m/x3c^2

The above is simplified to

T3 = (1)(a) + (1)(9.8)

T3 = 9.8 + a --- call this Equation 8

NOTE that since it was assumed that the pulley in the system is frictionless, then

T2 = T3

OR

Equation 7 = Equation 8, hence

Mb(g)[sin 37 - (mu)(cos 37)] + Mr(g)[sin 37 - (mu)(cos 37)] - a(Mb + Mr) = 9.8 + a

Solving for "a",

a + a(Mb + Mr) = Mb(g)[sin 37 - (mu)(cos 37)] + Mr(g)[sin 37 - (mu)(cos 37)] - 9.8

Substituting appropriate values and simplifying,

a + 7a = 20.5275

8a = 20.5275

a = 2.56 m/sec^2 --- this is the acceleration of the system

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The tensions on the strings are calculated as follows:

T3 = 9.8 + a (from Equation 8)

T3 = 9.8 + 2.56 = 12.36 N

and since T2 = T3,

T2 = 12.36 N

From Equation 4,

T1 = Mb(g)[sin 37 - (mu)(cos 37)] - Mb(a)

and substituting appropriate values,

T1 = 2(9.8)[sin 37 - (0.2)(cos 37)] - (2)(2.56)

T1 = 8.665 - 5.12

T1 = 2.545 N

ANSWER:

The system's acceleration = 2.56 m/sec^2

and the tensions in the strings are

T1 = 2.545 N

T2 = T3 = 12.36 N

NOTE --- I am hoping that my algebraic manipulations of the derivations and the subsequent arithmetic operations are correct. If, however, there have been algebraic and arithmetic errors, the analysis remains the same though.

Good luck.