Physics, tension in the guy wire?

Question

If the weight of a uniform boom is 2.00 kN, the tension in the horizontal guy wire supporting the boom a 5.60 kN weight is....?
Also given the length of the boom is 10 m and the height of the guy wire from the horizontal is 4 m.

Any help is appreciated - I have absolutely no idea what to do with this!

Answer 1

When you're dealing with an object that's in static equilibrium, you know that two thing must be true:

1) The sum of all the FORCES on the object is zero (true because it's not accelerating);
2) The sum of all the TORQUES on the object is zero (true because it's not rotating);

In some problems it's useful to analyze the forces; but in this problem it's probably more useful to analyze the torques.

When analyzing torques, you have to designate some point as the "rotation point" or axis -- but if it's not actually rotating, that choice is arbitrary. So let us aribitrarily choose the point where the boom is connected to the ground, and call that the "axis of rotation."

Now consider all the torques. Remember that:
torque = force × lever-arm
and remember that the lever arm is the _perpendicular_ distance between the line of force and the axis of rotation.

So, let's say the boom is extended diagonally upward to the right. In that case, we have three torques:

Torque 1 is the torque due to the hanging weight. The force is 5.60 kN. Since the line of force is vertical the lever arm is the _horizontal_ distance between the weight and the rotation point. From the pythagorean theorem that distance is: sqrt((10m)² − (4m)²) ≈ 9.2m.

So:
τ1 = (5.6 kN)(9.2m) [clockwise]

Torque 2 is due to the weight of the boom itself. In this case the line of force can be viewed as acting through the boom's center of mass, which is right in the middle of the boom. Therefore, the lever arm for Torque 2 is 1/2 of what the previous lever arm was:

τ2 = (2.00kN)(4.6m) [clockwise]

Torque 3 is due to the pull of the guy wire. Because it's pulling to the left, this torque is counter-clockwise. The force is the unknown tension T; and the lever arm is the _vertical_ distance from the guy wire to the rotation axis, which the problem says is 4m. So:

τ3 = (T)(4m) [counter-clockwise]

Now, there is one more force acting on the boom, and that's at the location where the boom is fastened to the ground. However, since that force passes _through_ the rotation axis, its lever arm is zero, so we don't have to count it in our inventory of torques. (That's one convenient reason for choosing that particular point as the rotation axis.)

So we have three torques, and we know that the "net" torque ought to come out to zero. That means the clockwise torques (τ1 and τ2) must exactly counterbalance the counter-clockwise torques (τ3):

τ1 + τ2 = τ3

(5.6 kN)(9.2m) + (2.00kN)(4.6m) = (T)(4m)

Now just solve that equation for "T".