1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F3=17 N. F4=15N and F5=9N 2)In the situation illustrated below, two blocks of Ma and Mb masses are suspended between two walls using ropes. If Ma=2kg, calculate the tension developed in the segments AB, BC and CD and the mass of block B for the equilibrium condition

3)In the figure below, the crane and its AB jib have together 390kg and center of mass in G1. If the 90kg BCD cage and the 80kg man have center of mass located in G2 and G3, respectively, (a) calculate the angle θ of inclination of the boom for which the crane is on the eminence of tumbling. (b) for which size the boom should be reduced (retracting its point B) so that it can be positioned horizontally (θ=0) and the crane does not tip over. Consider that the cage always stays in the horizontal position

4) A 1.7kg plate with center of mass in G is supported by a bar and three cables, as shown in the figure below. Determine the voltage developed in the AB, AC and DE cables and the reaction (there is only one) at point O for the equilibrium condition

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