Question

If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.

Answer 1

F_ab = 260.17 N

Step-by-step explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

Point A = 0.2 k

Point B = 0.4 i + 0.2 j

Point C = 0.2 j + 0.4 k

Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)

= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

vec(CD) = D - C = 0.4 i - 0.2 j

unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

= [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

(M)_c = ( r_ac x vec(F_ab) ) :

`\left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]`

(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD) is as follows:

(M)_cd = unit(CD) . (M)_c :

(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)

(M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

57 = F_ab*(sqrt(30) / 25)

F_ab = 260.17 N