For a rigid slab in translation, show that the system of the effective forces consists of vectors (Δmi)a¯ attached to the various particles of the slab, where 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

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.

Accepted Answer

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

The sum of these vectors is: [latex]\sum \left( \Delta { m }_{ i } ight) \overline { a }=\left( \sum\Delta { m }_{ i } ight) \overline { a }[/latex]

or since [latex]\sum\Delta { m }_{ i }=m[/latex] ,

[latex]\sum \left( \Delta { m }_{ i } ight) \overline { a }=m \overline { a }[/latex]

The sum of the moments about G is: [latex]\sum { { r }_{ i }^{ \prime } } imes \left( \Delta mi ight) \overline { a } =\left( \sum { \Delta { m }_{ i } } { r }_{ i }^{ \prime } ight) imes \overline { a }[/latex]

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

Question : For a rigid slab in translation, show that the system of the...

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