## Question:

For a rigid slab in translation, show that the system of the effective forces consists of vectors $\left( \Delta { m }_{ i } \right) \overline { a }$ attached to the various particles of the slab, where $\overline { a }$ is the acceleration of the mass center G of the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a single vector $m\overline { a }$ attached at G.

## Step-by-step

Since slab is in translation, each particle has same acceleration as G, namely $\overline { a }$ The effective forces consist of $\left( \Delta { m }_{ i } \right) \overline { a }$ .

The sum of these vectors is: $\sum \left( \Delta { m }_{ i } \right) \overline { a }=\left( \sum\Delta { m }_{ i } \right) \overline { a }$

or since $\sum\Delta { m }_{ i }=m$ ,

$\sum \left( \Delta { m }_{ i } \right) \overline { a }=m \overline { a }$

The sum of the moments about G is: $\sum { { r }_{ i }^{ \prime } } \times \left( \Delta mi \right) \overline { a } =\left( \sum { \Delta { m }_{ i } } { r }_{ i }^{ \prime } \right) \times \overline { a }$

But,$\sum { \Delta { m }_{ i } } { r }_{ i }^{ \prime }=m{ { r }^{ \prime } } =0$ because G is the mass center. It follows that the right-hand member of Eq. (1) is zero. Thus, the moment about G of $m\overline { a }$ must also be zero, which means that its line of action passes through G and that it may be attached at G.