Explanation:
1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis. The angle between F₁ and the +z axis is 30°. Therefore, the vector is:
<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)
<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)
<F₁> = 5 i + 5√3 j + 10√3 k
F₂ is in the xy plane. Its slope is -24/7. The vector is:
<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)
<F₂> = -3.36 i + 11.52 j
F₃ is parallel to the +x axis. The vector is:
<F₃> = 17 (i + 0 j + 0 k)
<F₃> = 17 i
F₄ is parallel to the -z axis. The vector is:
<F₄> = 15 (0 i + 0 j − k)
<F₄> = -15 k
F₅ is in the xy plane. It forms a 15° angle with the -y axis. The vector is:
<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)
<F₅> = -9 sin 15° i − 9 cos 15° j
The resultant vector is therefore:
<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k
<F> = 16.31 i + 11.49 j + 2.32 k
2) Sum of forces at point B in the x direction:
∑F = ma
Tbc cos 40° − ¹⁵/₁₇ Tab = 0
Tbc cos 40° = ¹⁵/₁₇ Tab
Tbc = 1.15 Tab
Sum of forces at point B in the y direction:
∑F = ma
Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0
Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)
(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N
1.21 Tab = 20 N
Tab = 16.52 N
Tbc = 19.02 N
Sum of forces at point C in the x direction:
∑F = ma
Tcd sin 25° − Tbc cos 40° = 0
Tcd sin 25° = Tbc cos 40°
Tcd = 1.81 Tbc
Tcd = 34.48 N
3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F. Relative to point A:
3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm
1336 kgm = 1530 kgm cos θ
θ = 29.17°
3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm
1336 kgm = (170 kg) x
x = 7.86 m
4) Find the lengths of the cables.
Lab = √((2 m)² + (3 m)² + (5 m)²)
Lab = √38 m
Lac = √((2 m)² + (3 m)² + (5 m)²)
Lac = √38 m
Lde = √((2 m)² + (3 m)²)
Lde = √13 m
Sum of forces in the x direction:
∑F = ma
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Sum of forces in the y direction:
∑F = ma
2/√38 Fab − 2/√38 Fac = 0
Fab = Fac
Sum of forces in the z direction:
∑F = ma
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
Sum of moments about the y-axis:
∑τ = Iα
(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0
Substitute Fab = Fac and simplify:
6/√38 Fab + 3/√13 Fde − mg = 0
30/√38 Fab + 6/√13 Fde − 2mg = 0
Double first equation:
12/√38 Fab + 6/√13 Fde − 2mg = 0
Subtract from the second equation:
28/√38 Fab = 0
Fab = 0
Fac = 0
Solve for Fde:
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
3/√13 Fde = mg
3/√13 Fde = (1.7 kg) (10 m/s²)
Fde = 20.43 N
Solve for Rx:
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Rx = 2/√13 Fde
Rx = 11.33 N
is the rest mass of the particle and c is the speed of light.
