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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.

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