Question

18. In Fig. 1.35, a uniform bar of length 1 m is supported
at its ends and loaded by a weight W kgf at its middle.
In equilibrium, find the reactions R, and R, at the ends.
RA
R2
W kgf
​

Answer 1

Answer:

R1+R2=w (Balancing Forces in Y−ax) __(I)

Writing equation for torque about the centre  of bar :

R1(0.5)+w(0)−R2(0.5)=0

R1=R2 __(II)

From eq. (I) & (II)

R1=R2=2wKgf.

Answer 2

According to the principle of moments,

Clockwise moments = Anticlockwise moments

R1 + R2 = W

Since the system is in equilibrium,

R1 × 1 / 2 = R2 × 1 / 2

R1 = R2

Therefore 2R1 = W

R1 = R2 = W / 2 kgf