The boom in the figure below ( Figure 1 ) weighs 2700 N and is attached to a fri

Question

The boom in the figure below (Figure 1) weighs 2700Nand is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 39.0% of its length.

A) Find the tension in the guy wire.

B) Find the horizontal component of the force exerted on the boom at its lower end.

C) Find the vertical component of the force exerted on the boom at its lowered.

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The boom in the figure below (Figure 1) weighs 2700Nand is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 39.0% of its length. Find the tension in the guy wire. Find the horizontal component of the force exerted on the boom at its lower end. Find the vertical component of the force exerted on the boom at its lowered.

Explanation / Answer

A) I assume the guy wire and the cable supporting the 5000 N mass are attached to the end of the boom and not rolling over pulleys.
I also assume that the guy wire is horizontal.

Let L be the boom length
Let T be the guy wire tension

Sum moments about the boom base pivot to zero. assume clockwise is positive moment.

0 = 5000(Lcos60) + 2700(0.39Lcos60) - TLsin60
as L is common in all terms, it will divide out.
0 = 5000(cos60) + 2700(0.39cos60) - Tsin60
T = 5000(cos60) + 2700(0.39cos60) / sin60
T = 3107.95 N

B) by summing forces in the horizontal direction to zero we see that the horizontal reaction must equal the tension in the guy wire
H = 3107.95 N

C) Sum forces in the vertical direction to zero.
0 = V - 5000 - 2700
V = 7700 N
Hope this helps :)

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