3 forces each equal to P act on a body, one at (a,0,0) parallel to OY. The second at the point (0,b,0) parallel to OZ and the third at the point (0,0,c) parallel to OX, the axis being rectangular.
1)Find the resultant force and the resultant moment
2) Find the pitch of the wrench
1) So if we take the forces parellel to OX , OY, and OZ axes respectively as F1,F2 and F3 (we know they have equal magnitudes P)
Resultant force, Fr = F1+F2+F3
They are in three planes
How do we sum them?
3 answers
Summing all three is a vector addition. Break up each force into three components (x,y,z). The resultant force will be the vector addition of each of these forces. Fr=F1 + F2 +F3 where each force has three directional components.
Since the forces F1, F2 & F3 passes through the points (a,0,0) , (0,b,0) & (0,0,c) and are parallel to OY, OZ, OX axes respectively,
can we take the sum as
F1=(j - ai)P
F2 =(k - bi)P
F3= (i - ck)P
F1+F2+F3 = [ (j - ai)P] + [ (k - bi)P ] + [ (i - ck)P ]= P[ (1-a)i + (1-b)j + (1-c)k ]
?
can we take the sum as
F1=(j - ai)P
F2 =(k - bi)P
F3= (i - ck)P
F1+F2+F3 = [ (j - ai)P] + [ (k - bi)P ] + [ (i - ck)P ]= P[ (1-a)i + (1-b)j + (1-c)k ]
?
yes, that is the approach.