The mobile crane has a weight of 127 kip and center of gravi

The mobile crane has a weight of 127 kip and center of gravity at G_1; the boom has a weight of 29 kip and center of gravity at G_2. Determine the smallest angle of tilt theta of the boom, without causing the crane to overturn if the suspended load is W = 42 kip. Neglect the thickness of the tracks at A and B. (Figure 1) Determine the smallest angle of tilt theta of the boom, without causing the crane to overturn. Express your answer to three significant figures and include the appropriate units.

Solution

The equation is really:
mass 1 x length 1 + mass 2 x length 2 = mass 3 x length 3

G1 * 6 ft = (G2 * X ft) + H3 * X ft)
G1 = 127 kip
G2 = 29 kip
H3 = 42 kip

Now it is an equilibrium of torque about the point where the boom pivots:

127 kip * 6 ft = 29 kip * Xft + 42 kip * Xft
solve for X feet:
X feet = 10.73 ft

So the crane is in equilibrium when the load is length1 + length 2 times the mass = mass of motor * length 3
(10.73 ft * 29 kip) + (10.73 ft * 42 kip) = 127 kip * 6 ft

The boom forms a triangle with the hypotenuse equals 12ft + 15 ft = 27 ft.
The bottom of the triangle is 10.73 ft + 10.73 ft = 21.46ft

So we have two sides of the triangle and now just need to find the angle where the triangle forms with those two sides:

cosine of angle = adjacent side / hypotenuse
cosine of angle = 21.46 / 27 = 0.80
and the arc cosine of that = 36.86 degrees

So rounding it off, angle of crane = 37 degree