Force, Position, Torque

Homework Torque 




First to determine the position vector (from the origin) for a point, you just need to take the projections of the vector (or of the ending point) on x, y, and z axis multiply them with the unity vectors i, j, k and add them algebraically.

For example for segment AC, A is the origin, the projections of C on x axis is 4 m, on y axis is 3 m and on z is 0 m.

R(AC) = 4*i + 3*j +0*k

For segment AB you either take the projections of B on the axes as above or simply sum

R(AB) = R(AC) + R(CB) = 4*i+3*j -2*k    where R(CB) = -2*k  (it has only a projection on z axis)
The unit vector for AC is simply R(AC)/module[R(AC)]

u(AC) = R(AC)/module[R(AC)] $= (4*i+3*j)/(sqrt{(4*4 +3*3)})$

Now the momentum of F about point C is by definition M(C) = R(CB) x F (vector product)

and the momentum of F about AC axis is the projection of the above momentum on the AC axis

M(AC) = u(AC)*M(C)   (scalar product)
M(AC) = u(AC)*(R(CB) x F)

Described different (but with same result) the momentum of F about the AC axis is the projection of the plane (xy) of the momentum of F about point A

M(A) = R(AB) x F

M(AC) = u(AC)*M(A) = u(AC)*(R(AB) x F)

Now to obtain the vector product of two vectors you need to compute the following determinant

x F =    i      j      k    =  i*(y*Fz-z*Fy) + j*(z*Fx – x*Fy) + k*(x*Fy – y*Fx)
               x     y      z
               Fx   Fy     Fz

Then to compute the scalar product of two vectors
u*M = ux*Mx + uy*My + Uz*Mz

Which will give you the following multiplication rule

u*(R x F) =  ux    uy    uz   = ux*(y*Fz – z*Fy) + ……
                   x      y      z
                   Fx    Fy     Fz

The last thing done on the paper is to express the vector M(AC) which module you computed (14.4 kN*m) by projecting its components to x and y directions. For this you just need to multiply the module of M(AC) with the components of the unity vector u(AC)
M(AC) = module(M(AC)) * u(AC) = 14.4*u(AC)