A mobile is a decorative structure that is suspended so as to tur
n freely in the air. A mobile usually consists of pieces of rods and decorative shapes made of metal, wood, plastic or paper suspended in midair by wires, strings or ties so that the individual parts can move independently when stirred by a breeze.
n freely in the air. A mobile usually consists of pieces of rods and decorative shapes made of metal, wood, plastic or paper suspended in midair by wires, strings or ties so that the individual parts can move independently when stirred by a breeze.
The principles in engineering mechanics about force systems in equilibrium can be demonstrated by analyzing the equilibrium condition of a mobile. Consider the fish mobile which hangs in our room. The fish mobile is an example of a parallel force system. All forces acting on the horizontal rods are vertical since only gravity loads due to the weights of the hanging objects act. I laid the mobile on a flat surface and this is how it will appear as a parallel force system.
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| The Fish Mobile as a Parallel Force System |
Haven’t you wondered how much force each wire supports to keep the mobile in equilibrium? Let’s apply the principles of engineering mechanics to answer this query. Here is model of the fish mobile as a parallel force system and the free body diagrams (FBD) of the rods supporting the fish mobile whrere the fish shapes are represented by their weights (W) and the tensile forces acting on the wires are represented as T1, T2, T3 and T4.
To compute the force acting on wire no. 4, the FBD with T4 is used and the equilibrium equation of summation of forces with respect to the vertical axis is applied. The upward forces should be equal to the downward forces. Hence, T4 = 2W. Similarly, the forces on the other wires can be computed by using the appropriate FBDs. Thus,
T3 = 3W. T2 = T4 + W = 2W + W = 3W.
T1 = T2 + W + T3 = 3W + W + 3W = 7W
The mass of each fish object is about 10 g or 0.01 kg which is equivalent to 0.0981 N. Hence, the tensile force acting on wire no. 1 is T1 = 7(0.0981 N) = 0.6867 N.
Except for second rod, all rods have symmetrical loading resulting to the wire located at the middle of the rod. However, for the rod where T2 acts, the force at the right is T4=2W and the force at left is W. For equilibrium, the location of the T2 is closer to right. An equation of equilibirum (summation of moments) may be used to determine the location of the wire for this rod.
The arrangement of the rods and hanging objects on a mobile will affect the magnitude of the forces acting on the wires. In the design of a mobile, the following factors should be considered: (a) Mass of the each hanging object and (b) Length of each rod. The location of the wire on the rod depends on the weights supported by the rod. If the weights are hanged symmetrically, then the wire is placed at the midspan of the rod to satisfy equilibrium. However, for unsymmetrical loading where the weight at the left is not equal to the weight at the right, the location can be obtained by applying the equation of equilibrium for moments.
Equilibrium of a fish mobile? That's simply mechanics.
Equilibrium of a fish mobile? That's simply mechanics.
