Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = 
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Here we deal with a lever law. It states that product of force and distance from a fixed point on a lever is equal on both sides.
F₁*d₁ = F₂*d₂
By analysing this formula we can see that applying small force on a great length equals great force on a small length.
To remove nail we need to apply certain force. If we use F₁ for this required force we can see that on other side we need to apply certain force. If we have greater arm length we need smaller force. In a crowbar arm length along which we apply force is greater than length of our arm. This leads to a conclusion that we need smaller force when using crowbar. Depending on the length of a nail it is possible that we need to apply force that is greater than force required to remove nail.
Density is given as
now we have to convert this density into 
now we have
We apply the following equation
T = 2π * sqrt (L/g)
Where g is the gravity = 9.8 m/s^2
L is the longitude of the pendulum (Height of the tower)
T is the period. (T = 18s)
We find L.............> (T /2π)^2 = L/g
L = g*(T /2π)^2...........> L = 80.428 meters
