Edexcel M5 2009 June — Question 5

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2009
SessionJune
TopicMoments

  1. Two forces \(\mathbf { F } _ { 1 } = ( 2 \mathbf { i } + \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 } = ( - 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }\) act on a rigid body. The force \(\mathbf { F } _ { 1 }\) acts at the point with position vector \(\mathbf { r } _ { 1 } = ( 3 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }\) and the force \(\mathbf { F } _ { 2 }\) acts at the point with position vector \(\mathbf { r } _ { 2 } = ( \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }\). A third force \(\mathbf { F } _ { 3 }\) acts on the body such that \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\) are in equilibrium.
    1. Find the magnitude of \(\mathbf { F } _ { 3 }\).
    2. Find a vector equation of the line of action of \(\mathbf { F } _ { 3 }\).
    The force \(\mathbf { F } _ { 3 }\) is replaced by a fourth force \(\mathbf { F } _ { 4 }\), acting through the origin \(O\), such that \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 4 }\) are equivalent to a couple.
  2. Find the magnitude of this couple.