In the previous parts, you obtained the following equations using Newton's 2nd law and the constraint on the motion of the two blocks: m2a2x=T−m2gsin(θ),(1) m1a1y=T−m1g,(2) and a2x=−a1y.(3) Solve these equations to find a1y. Before you enter your answer, make sure it satisfies the special cases you already identified: a1y=−g if m2=0 and a1y=0 if m1=m2 and θ=π/2. Also make sure that your answer has dimensions of acceleration. Express a1y in terms of some or all of the variables m1, m2, θ, and g.

1)

2) See below

Explanation:

1)

The system of equations that we have in this problem is:

(1)

(2)

(3)

Here we want to solve the system for .

First of all, we isolate T from eq.(2):

And we substitute this expression into eq(1):

Solving for , we get

Now we can substitute this into eq(3):

And re-arranging for we find:

2)

We see that this solution satisfies all special cases. In fact:

- If ,

- If and , we get

Also, this solution has dimensions of acceleration. In fact:

- The term at the numerator is in the form , where mass is in kilograms and g is in meters per second squared (so, it is in the form mass*acceleration)

- The term at the denominator is a mass, so it is in kilograms

This means that is (mass*acceleration)/(mass), therefore it has dimensions of acceleration.

trig 15.3 cos