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J - Moment about an Axis Angel

# J - Moment about an Axis Angel - FORCE SYSTEM RESULTANTS...

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Unformatted text preview: FORCE SYSTEM RESULTANTS MOMENT ABOUT AN AXIS LEARNING OBJECTIVES Be able to determine the moment of a force about a specified axis using scalar and vector methods PRE-REQUISITE KNOWLEDGE Units of measurements Trigonometry concepts Vector concepts Rectangular components concepts Moment concept Cross product and dot product MOMENT OF A FORCE ABOUT A SPECIFIED AXIS (Scalar Method) M ob = F db Where Mob is the moment of the force F about the Ob axis and db is the perpendicular or the shortest distance from the force to the axis EXAMPLE The Moment arm about the y-axis is 0.3 m, My = 6 N-m The Moment arm about the Ob-axis is 0.5 m, Mob = 10 N-m The Moment arm about the x-axis, is 0.4 m, Mx = -8 N-m MOMENT OF A FORCE ABOUT A SPECIFIED AXIS (Vector Method) M OB = rOA F ( M Ob = u Ob rOA F ( ) ) Where uOb is the unit vector in the direction of the Ob-axis M Ob i = ( u bx i + u by j + u bz k ) rx Fx j ry Fy k u bx rz = r x Fz Fx u by ry Fy u bz rz Fz EXAMPLE 1 rOb =- i +3 j 4 u Ob u Ob rOb - 4i + 3 j = = 2 2 rOb ( 4) + 3 -4 3 = i+ j 5 5 M Ob = u Ob ( r F ) M Ob uobx = rX FX uoby rY FY uobz rZ FZ 4 5 = 0.3 0 3 5 0.4 0 0 0 - 20 EXAMPLE 1 M Ob 4 5 = 0.3 0 3 5 0.4 0 0 0 - 20 M Ob 4 3 = - ( 0.4 -20 - 0 0 ) - ( 0.3 -20 - 0 0 ) = 10 N - m 5 5 EXAMPLE 2 Determine the moment of the force F about the Oa axis using vector method SOLUTION 1 r ar Determine the directional and the unit vectors in the direction of the Oa axis : rOa = 4 j + 3k u Oa rOa = rOa Determine the moment arm vector u Oa 4 3 = = j+ k 2 2 5 5 4 +3 4 j + 3k m rarm = i - 2 j + 6k SOLUTION 1 continued Determine the moment about the Oa axis M Oa = u Oa (rarm F ) M Oa ux = rx Fx 0 = 1 50 uy ry Fy 4 5 -2 - 20 uz rz Fz 3 5 6 20 rarm 4 3 = 0[(-2)(20) - (-20)(6)] - [(1)(20) - (6)(50)] + [(1)(-20) - (-2)(50)] 5 5 = 0 + 224 + 48 = 272 N - m SOLUTION 1 continued rarm In vector form Moa can be obtained as Moa(uoa) M Oa M Oa 4 3 = u Oa ( M oa ) = 272 0i + j + k 5 5 = 217.6 j + 163.2k EXAMPLE 2 Determine the moment of the given 20 lb force about the hinge using vector method SOLUTION 2 Determine the unit directional vector in the direction of hinge : uhinge u hinge = i Determine the force vector using the following 3 steps: 1. Find the position vector rBA 2. Find the unit vector uBA 3. Find the force vector F = FuBA rBA = 3i - 3 cos 20 j + ( 4 - 3 sin 20) k = 3i - 2.819 j + 2.974k u BA = 3i - 2.819 j + 2.974k 3 + ( - 2.819 ) + 2.974 2 2 2 = 0.5907i - 0.5551j + 0.5856k F = F u BA = ( 20)( 0.5907i - 0.5551j + 0.5856k ) F =11.81i -11.10 j +11.71k lb SOLUTION 2 continued Determine the moment arm or the position vector rOB uhinge rOB = 3 cos 20 j + 3 sin 20k rOB Determine the moment at the hinge Mhinge M hinge = u hinge (rarm F) = ux rx Fx uy ry Fy uz rz = Fz 1 0 11.81 0 0 3 cos 20 3 sin 20 = 44.4 lb - ft - 11.10 11.71 ALTERNATIVE SOLUTION 2 Determine the unit vector in the direction of hinge : uhinge u hinge =i As before, the force vector F is F = {11.81i -11.10 j +11.71k} lb ALTERNATIVE SOLUTION 2 continued Determine the position vector rarm from O to A rarm rarm = rOA = 3i + 4k Determine Mhinge M hinge = u hinge (rarm F ) = ux rx Fx uy ry Fy uz 1 0 0 rz = 3 0 4 = 44.4 lb - ft Fz 11.81 -11.10 11.71 ...
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