Question

Learning Goal: To gain insight into the independence of the scalar triple product from the point on the line chosen as the reference point of the calculation.

Answer 1

Given that :

dimensions are at x₁ = 1.4 m,   y₁ = 1.7 m & z₁ = 1.5 m

applied force at point C, F = | -165 i + 100 j + 140 k | N

Part-B : the moment about AB using the position vector AC which is given as -

using an equation, M_AB = u_AB . (r_AC x F)                                                                   { eq.1 }

we have OA = 0 , OB = 1.4 i + 1.7 j and   OC = 1.4 i + 1.7 j + 1.5 z

find out, vec/AB = vec/OB - vec/OA

vec/AB = (1.4 i + 1.7 j ) - (0)

vec/AB = 1.4 i + 1.7 j

And   u_AB = vec/AB / | vec/AB |                                                                       { eq.2 }

| vec/AB | = magnitude of AB = sqrt [(1.4)² + (1.7)²

inserting the value in eq.2,

u_AB = (1.4 i + 1.7 j) / sqrt [(1.4)² + (1.7)²

u_AB = (1.4 i + 1.7 j) / (2.20)

u_AB = (0.636) i + (0.772) j

distance, r_AC = vec/OC - vec/OA

r_AC = (1.4 i + 1.7 j + 1.5 z)

inserting the value of u_AB , r_AC & F in eq.1,

M_AB = u_AB . (r_AC x F) = | $\hat{i}$    _$\hat{j}$        $\hat{k}$ |

                                     | 0.636       0.772              0 |

                                     | 1.4            1.7            1.5 |

                                     | -165         100             140 |

M_AB = (0.636) [1.7 x 140 - 1.5 x 100] + (0.772) [1.4 x 140 + 165 x 1.5] + (0)

M_AB = (0.636) (238 - 150) + (0.772) (196 + 247.5) + 0

M_AB = (55.9) $\hat{i}$ + (342.3) _$\hat{j}$ + (0) $\hat{k}$ N.m

Part-C : the moment about AB using the position vector BC which is given as -

using an eq.1, M_AB = u_AB . (r_BC x F)      

where, r_BC = vec/OC - vec/OB

distance, r_BC = (1.4 i + 1.7 j + 1.5 z) - (1.4 i + 1.7 j)          

r_BC = 1.5 z

inserting the value of u_AB , r_AC & F in eq.1,                    

M_AB = u_AB . (r_BC x F) = | $\hat{i}$    _$\hat{j}$    $\hat{k}$ |

                                     | 0.636       0.772             0 |

                                     | 0               0        1.5 |

                                     | -165         100             140 |

M_AB = (0.636) [0 x 140 - 1.5 x 100] + (0.772) [0 x 140 + 165 x 1.5] + (0)

M_AB = (0.636) (0 - 150) + (0.772) (0 + 247.5) + 0

M_AB = - (95.4) $\hat{i}$ + (191.07) _$\hat{j}$ + (0) $\hat{k}$ N.m