A counterweighted security barrier arm is designed to be assisted in opening by a falling weight, as shown in the diagram below.
Weight A is a solid metal cylinder with the following properties:
- diameter = 6.0×10−2 m6.0 \times 10^{-2}\text{ m}6.0×10−2 m
- length = 0.35 m0.35\text{ m}0.35 m
- weight = 44 N44\text{ N}44 N
Table 1 below lists the density of three candidate metals.
Table 1
| Material | Density / kg m−3\text{kg m}^{-3}kg m−3 |
|---|---|
| Aluminium | 2.7×1032.7 \times 10^32.7×103 |
| Titanium | 4.5×1034.5 \times 10^34.5×103 |
| Zinc | 7.1×1037.1 \times 10^37.1×103 |
Deduce which one of the three metals is used for weight A.
All pulleys are frictionless, so the tension in the rope supporting weight A is equal to its weight. The weight of the movable pulley M is negligible.
Calculate the tension in the horizontal cable C when the barrier is in the position shown above.
As the barrier arm is lowered, the movable pulley M is pulled to the left. Explain why this increases the tension in the horizontal cable C.
The side-view diagram above shows the barrier arm when it is partially raised. The horizontal cable C passes over a small pulley R and is attached to the arm at point D.
- The angle between the barrier arm and the horizontal cable C is 22∘22^\circ22∘.
- The distance between the pivot and D is 1.2 m1.2\text{ m}1.2 m.
- The tension in the horizontal cable C in this position is 65 N65\text{ N}65 N.
Calculate the moment of the tension about the pivot.
Suppose this counterweight system is attached to a heavier barrier arm that requires a larger moment about the pivot to raise it.
Identify and explain two independent design modifications to the system that would increase the moment about the pivot exerted by the horizontal cable C.