Example

Question:

A metal bar 70 cm long and 4.00 kg in mass is supported on two knife-edges placed 10 cm from each end.
A 6.00 kg load is suspended at 30 cm from one end.
Find the reactions at the knife-edges.
(Assume the bar to be of uniform cross section and homogeneous.)

Solution:

Length of bar, \( AB = 70\,\mathrm{cm} \)
Knife edges at \( K_1 \) and \( K_2 \), both 10 cm from ends.
Mass of bar, \( M = 4.00\,\mathrm{kg} \); loaded mass \( m = 6.00\,\mathrm{kg} \) at point \( P \) (=30 cm from end A).
Centre of gravity (\( G \)) at centre (\( AG=35\,\mathrm{cm} \)), and \( AP=30\,\mathrm{cm} \), \( PG=5\,\mathrm{cm} \), \( K_1G=25\,\mathrm{cm} \), \( K_2G=25\,\mathrm{cm} \).

The weights:
\[ W = 4.00g,\quad W_1 = 6.00g \] where \( g = 9.8\,\mathrm{m/s}^2 \).
For translational equilibrium:
\[ R_1 + R_2 - W_1 - W = 0 \implies R_1 + R_2 = (4.00 + 6.00) \times 9.8 = 98.00\,\mathrm{N} \]

For rotational equilibrium about \( G \):
\[ -R_1(K_1G) + W_1(PG) + R_2(K_2G) = 0 \] Substitute values:
\[ -R_1 \times 0.25 + 6.00g \times 0.05 + R_2 \times 0.25 = 0 \] \[ -0.25 R_1 + 2.94 + 0.25 R_2 = 0 \] \[ R_1 - R_2 = 1.2g = 11.76\,\mathrm{N} \]

Solve equations:

• \( R_1 + R_2 = 98.00\,\mathrm{N} \)
• \( R_1 - R_2 = 11.76\,\mathrm{N} \)

Adding: \[ 2R_1 = 109.76 \implies R_1 = 54.88\,\mathrm{N} \] Subtracting: \[ 2R_2 = 86.24 \implies R_2 = 43.12\,\mathrm{N} \]
Thus, the reactions at the supports are about \( 55\,\mathrm{N} \) at \( K_1 \) and \( 43\,\mathrm{N} \) at \( K_2 \).