ME 2560 Statics Moment of a Force about an Axis Two Dimensional Systems o The moment of a force about a point O may be calculated using a cross product. MO r F o Here, M O is perpendicular to the plane formed by the vectors r and F , and it has magnitude | M O | | F | d | F | r sin( ) . o In the two dimensional system shown, M O represents the moment of the force about an axis perpendicular to the page (in the k direction) and passing through O. A O θ θ Line of Action Three Dimensional Systems o To find the moment of a force about an axis in three dimensional analysis, we first calculate the moment about any point on that axis, say O, then we project that moment onto the axis using the dot product. M n -axis M O n r F n o Here, M n -axis is the scalar moment of F about the axis passing through O and parallel to the unit vector n . In vector form, we write M n-axis M O n n o M n -axis can be positive or negative depending on the angle between n and M O . If it is positive, point your right thumb in the direction of n and your right fingers will show the circulation of F about the axis. If it is negative, point your right thumb opposite the direction of n and your right fingers will show the circulation of F about the axis. o As before, r is a position vector from O to any point on the line of action of F . Kamman – ME 2560: page 1/2 Example: Given: F 100 i 50 j 200 k (lb) ; A : (3,4,5) (ft) Find: a) M x , M y , and M z the moments of F about the X, Y, and Z axes, b) M n -axis the scalar moment of F about an axis in the X-Y plane that makes an angle of 30 (deg) with the X-axis, and c) M n -axis Solution: a) M O r A/O F i j k 3 4 5 800 250 i 600 500 j 150 400 k 100 50 200 550 i 1100 j 550 k (ft-lb) M x M O i 550 (ft-lb) , M y M O j 1100 (ft-lb) , M z M O k 550 (ft-lb) b) n cos(30) i sin(30) j M n-axis M O n 550cos(30) 1100 sin(30) 550 0 73.686 73.7 (ft-lb) c) M n -axis M O n n 73.686 n 63.8 i 36.8 j (ft-lb) Note on Calculation of the Scalar Moment: M n -axis o Calculation of the scalar moment can also be done in determinant form. Simply replace the first row of the determinant by the components of n and expand as usual. nx ny nz M n -axis rx ry rz nx ry Fz rz Fy n y rx Fz rz Fx nz rx Fy ry Fx Fx Fy Fz o So, for the above example, we have cos(30) sin(30) M n -axis 3 100 4 50 0 5 cos(30) 800 250 sin(30) 600 500 200 73.7 (ft-lb) Kamman – ME 2560: page 2/2
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