PX155 - B3 - forces out of balance

PX155 - A6 - block on a slope
- but now, the block is sliding $\to$ friction is kinetic
  - $a$ is not a force, so indicated by a double header arrow
- by PX155 - A2 - newton's second law
  - parallel to the slope: $W \sin θ - F_{k} = m a$
  - perpendicular the the slope: $W \cos θ = 0$
    where, $W = m g$ ; $N = m g \cos θ$
    $F_{k} = μ_{k} N = μ_{k} m g \cos θ$
    $m a = m g \sin θ - μ_{k} m g \cos θ$
    $a = g \sin θ - μ_{k} g \cos θ = g (\sin θ - μ_{k} \cos θ)$
    $a$ is constant
eg: reconsider the block on a slope, but now, the slope is to the upper surface of a wedge and the wedge is resting on a table and there is no friction. to find: accelerations of the block and the wedge
- draw two diagrams, one for each body
- ${\vec{a}}_{1}$ is not the actual acceleration of the block (relative to the table), it is the acceleration relative to the wedge, which doesn't define the initial reference frame
- we need to apply PX155 - A2 - newton's second law in the frame defined by the fixed reference

{\vec{a}}_{B} = {\vec{a}}_{1} + {\vec{a}}_{2}

- for wedge, horizontally:

N \sin θ = m_{W} a_{2}

- for block, perpendicular to the slope:

N - m_{B} g \cos θ = - m_{B} a_{2} \sin θ

- combining above equations:

\begin{array}{r} m (g \cos θ - a_{2} \sin θ) \sin θ = m_{W} a_{2} \\ a_{2} = \frac{m_{B} g \cos θ \sin θ}{m_{W} + m_{B} \sin^{2} θ} \end{array}

- for block, parallel to the slope:

m g \sin θ = m (a_{1} - a_{2} \cos θ)

- combining equations above:

a_{1} = g \sin θ + \frac{m_{B} g \cos^{2} θ \sin θ}{m_{W} + m_{B} \sin^{2} θ}