Equilibrium of a Rigid Body in Space

Equilibrium of a Rigid Body in Space

• For a non-concurrent general force system in equilibrium, the resultant force and the resultant moment about any point is zero.

• As moment about a point is a vector in three dimensions, we get three equations by equating moment to zero.

• Thus six equations can be obtained which can be solved for six unknowns.

• For equilibrium of parallel force system, one of the equations can be obtained by equating summation of all forces, along an axis parallel to the forces to zero.

• Additional equations can be obtained by equating moments of all forces about different lines in a plane perpendicular to the forces to zero.

Solved Examples for Understanding

Example 4.8.1 

Determine tensions is cords BD and BE which hold a rigid light member in horizontal position as shown. Assume A as ball and socket joint.

Solution: 

The unknown reactions at A can be avoided in equations by taking moment about A.

Example 4.8.2 

A plate ACED weighs 7,600 kg/m3. It is held in horizontal plane by three wires at A B and C. Find tensions in the wires. Refer Fig. 4.8.2.

Solution: 

The weight of the plate is

W = ρ × g × Volume

= 7600 × 9.81 × 0.9 × 1.2 × 10 × 10-3

⸫ W = 805.205 N

It acts through the C.G. shown in Fig. 4.8.2 (a).

Example 4.8.3

Fig. 4.8.3 shows steel pipe ABCD of length 1.8 m having 90° bend at B. The pipe weights 30 N/m and supported by three vertical wires attached to the pipe at A, C and D. Find tension in wires.

Solution: 

The weights of AB and CD are shown acting through their centres in Fig. 4.8.3 (a).

Consider moments about a line DE parallel to z-axis.

Example 4.8.4 

The uniform concrete slab has a weight of 5500 N. Determine tension in each of the three parallel supporting cables when the slab is held in horizontal plane as shown in Fig. 4.8.4.

Solution: 

Weight of slab acts at its centre

⸫ (TB) (0.6) + (5500) (0.3) = 0