Question

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.

Answer 1

Answer:

1) $a_{1y}=\frac{m_2 g sin \theta - m_1 g}{m_1 + m_2}$

2) See below

Explanation:

1)

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

$m_2 a_{2x}=T-m_2 g sin \theta$ (1)

$m_1 a_{1y}=T-m_1 g$ (2)

$a_{2x}=-a_{1y}$ (3)

Here we want to solve the system for $a_{1y}$.

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

$T=m_1 a_{1y}+m_1 g$

And we substitute this expression into eq(1):

$m_2 a_{2x}=(m_1 a_{1y}+m_1 g)-m_2 g sin \theta$

Solving for $a_{2x}$, we get

$a_{2x}=\frac{(m_1 a_{1y}+m_1 g)-m_2 g sin \theta}{m_2}$

Now we can substitute this into eq(3):

$\frac{(m_1 a_{1y}+m_1 g)-m_2 g sin \theta}{m_2}=-a_{1y}$

And re-arranging for $a_{1y}$ we find:

$(m_1 a_{1y}+m_1 g)-m_2 g sin \theta}=-m_2 a_{1y}\\a_{1y}(m_1+m_2)=m_2 g sin \theta - m_1 g\\a_{1y}=\frac{m_2 g sin \theta - m_1 g}{m_1 + m_2}$

2)

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

- If $m_2=0$, $a_{1y}=\frac{0\cdot g sin \theta - m_1 g}{m_1 + 0}=\frac{-m_1 g}{m_1}=-g$

- If $m_1 = m_2$ and $\theta=\frac{\pi}{2}$, we get

$a_{1y}=\frac{m_2 g sin \frac{\pi}{2} - m_2 g}{m_2 + m_2}=\frac{m_2 g-m_2 g}{2m_2}=0$

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

- The term at the numerator is in the form $m*g$, 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 $a_{1y}$ is (mass*acceleration)/(mass), therefore it has dimensions of acceleration.