Question

<Extra Credit 2 Force Vector Directed along a Line Three-dimensional Cartesian force vectors are used throughout engineering mechanics. The generic force vector is represented as follows: F= F,i+Fyj + F kwhere F is the force vector and Fr, Fy, and F, are the vector's i, j, and k components, respectively. The force vector has a magnitude F F2 +F3 + F2. The vector's I 1 Figure < 1 of 2 > C 13/12 5 F = clb b ft у A aft

Solve all parts

thanks

Answer 1

Solution: 1312 tano = 12 5 5 12 = 67.38 cc tan 67-38 co oc 67-38° 4ft 4ft Tel = 9.644 OA = 631 to to t 41 oc - – 49 +9.6k AC oc - o (-49 + 9.6 h) – (6) lado -42-63 1962 (AC) = (-4) + (-63² + (9.63² SAC) = 12 - 4 - 69 +9.6 unit vector along AC 19 . - - 4 5 1. -0.3331 -0.5 + 0.8 +0.8k

AC 74 = 30 (-0.333? – 0.59 +0.8k) - 0.59 + 0.8k) 101 - 159+2 = 30 (-17 3 1 JI + 24k F = - 10 lb = 15 lb +24 lb Fy Ez = Part Bi OR = x + y + z lonla x² + y² + 2? e Ej j & Fak F'! 2 F + Fyr + F DA This is unit vector LOA FI

ОА (Er it Ey 3 + E k) a 2 Ey + F (Fx il + Ey ) + Ez k) X x + y ) + k afa + aty } + atz Х Х = a Fc Х عا a Fy X z afz X

At first inclination of OC is found. From that distance along positive z-axis is found.

Then vector AC is determined from OC and OA. After that unit vector along AC is found.

Then multiplying 30 with the unit vector along AC gives us the force vector along AC. From that force components are noted down.

For part B, the key step is to find unit vectors for both position of A and the force direction.

i.e. first find vector OA and unit vector along the direction.

And then from force vector find another unit vector.

Now, both the unit vectors are the one and the same. By equating them we can find x,y, and z components of A in a, Fx , Fy , Fz and X.

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