A100 kg nonuniform boom, 6. 0 m long, is loosely pinned at the pivot at P. A 700
ID: 1490585 • Letter: A
Question
A100 kg nonuniform boom, 6. 0 m long, is loosely pinned at the pivot at P. A 700 kg block is suspended from the end of the boom at A. The boom forms a 30degree angle with the horizontal, and is supported by a cable, 4. 0 m long, between points D and B. Point B is 4. 0 m from P, and point D is 4. 0 m above P. The center of mass of the boom is at point C, which is 2. 0 m from P. In the figure, the y-component of the pivot force on the boom at P is closest to: A) 2500 N B) 8100 N C)6100 N D)7800 N E)4200 NExplanation / Answer
Because the center of gravity of the boom is equidistant from the pivot and the attachment of the cable at point B, the 980 Newton weight of the boom is borne equally by the pivot and the cable.
The 700 newton weight is 50% further from the wall than the cable attachment at point B, therefore the cable will bare 150% of the object's weight.
The vertical component of the tension on the cable is 10,780 Newtons.
100 * 9.8 / 2 + 700 * 9.8 * 1.5 = 10,780
The horizontal component of the tension is 18,671.508 Newtons.
10,780 / tan(30) = 18,671.508
This is also the horizontal component (though opposite) of the force of the pivot on the boom.
The total tension of the cable is 21,560 Newtons.
10,780 / sin(30) = 21,560
The vertical component of the force if the pivot on the boom is the arithmetic difference between the vertical component of the cable's tension and the total weight of the system.
10,780 - (100 + 700) * 9.8 = 2,940 Newtons. closest is 2500 N
Notice that the vertical component of the tension is greater than the total weight of the system.
This is because the geometry causes the boom, which acts as a lever, to push upward on the pivot.
...and the pivot, correspondingly, pushes down on the boom.
This downward force that the pivot applies to the boom is then borne by the cable.
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