Wat formulas should i use with this incline?

Question

A block (mass m1) lying on a frictionless inclined plane is connected to a mass m2 by a massless cord passing over a pulley.

A) determine a formula for the acceleration of the system of the two blocks in terms of m1, m2, Θ and g.

b) what conditions apply to masses m1 and m2 for the acceleration to be in one direction (say, m1 down the plane), or in the opposite direction?

Answer 1

If you draw the free body diagram of the system, you will note the following forces acting on "m1", and using Newton's 2nd Law of Motion,

T1 - m1(g)(sin Θ) = m1(a)

where

T1 = tension in the cord pulling the body "m1"

and all the rest of the variables are given in the problem.

Solving for T1,

T1 = m1(a) + m1(g)(sin Θ) --- call this Equation 1

The following forces are acting on "m2" and again using Newton's 2nd Law of Motion,

m2(g) - T2 = m2(a)

where

T2 = tension in the cord holding the body "m2"

Solving for T2,

T2 = m2(g) - m2(a) --- call this Equation 2

The pulley in this particular problem will be ASSUMED to be frictionless as well. Thus being said, the tension in the massless cord is constant. Hence,

T1 = T2  -------------- Equation 1 = Equation 2

m1(a) + m1(g)(sin Θ) = m2(g) - m2(a)

Rearranging the above,

m1(a) + m2(a) = m2(g) - m1(g)(sin Θ)

Factoring out "a" and "g",

a(m1 + m2) = g(m2 - m1*sin Θ)

and solving for "a",

a = [g(m2 - m1*sin Θ)]/(m1 + m2)

The above derived formula assumes that "m1" is being pulled by "m2", hence, the motion can be described as "m1" going up the plane and, obviously, "m2" is going down.

The condition of the system's direction is determined by the factor

(m2 - m1*sin Θ)

and as long as

(m2 - m1*sin Θ) > 0, then the motion is that as described above. However, if

(m2 - m1*sin Θ) < 0, then the system will be moving in the opposite direction, i.e., m1 will be going down the plane and m2 going up towards the pulley.

Hope this derivation and analysis will help you out. Good luck.