The mobile crane has a weight of 116kip and center of gravity at G1; the boom has a weight of 29kip and center of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn if the suspended load is W = 38kip . Neglect the thickness of the tracks at A and B. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn.
The key here is to simplify this as much as possible 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 = 116 kip G2 = 29 kip H3 = 38 kip Now it is an equilibrium of torque about the point where the boom pivots: 116 kip * 6 ft = 29 kip * Xft + 38 kip * Xft solve for X feet: X feet = 10.388 ft So the crane is in equilibrium when the load is length1 + length 2 times the mass = mass of motor * length 3 (10.388 ft * 29 kip) + (10.388 ft * 38 kip) = 116 kip * 6 ft The boom forms a triangle with the hypotenuse equals 12ft + 15 ft = 27 ft. The bottom of the triangle is 10.388 ft + 10.388 ft = 20.776ft 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 = 20.766 / 27 = 0.7691 and the arc cosine of that = 39.7 degrees So rounding it off, angle of crane = 40 degree
