A Classical Approach of Newtonian Mechanics

10 STATICS 10.6 Jointed rods

N

Setting the net horizontal force on the rod to zero gives

Y 1 + Y 3 = 0.

Setting the net vertical force on the strut to zero yields

X 2 − X 3 = 0.

Finally, setting the net horizontal force on the strut to zero yields

Y 2 − Y 3 = 0.

The above equations can be solved to give

−Y 1 = Y 2 = Y 3 = Y,

and

with

X 2 = X 3 = X,

X 1 = m g − X.

There now remain only two unknowns, X and Y.

Setting the net torque acting on the rod about the point where it is connected

to the wall to zero, we obtain

m g l/2 − X 3 l = 0,

where l is the length of the rod. Here, we have used the fact that the centre of
gravity of the rod lies at its mid-point. The above equation implies that

X 3 = X = m g/2 = 15 × 9.81/2 = 73.58 N.

We also have X 1 = m g − X = 73.58 N. Setting the net torque acting on the strut
about the point where it is connected to the wall to zero, we find

Y 3 h sin θ − X 3 h cos θ = 0,

where h is the length of the strut. Thus,

Y 3 = Y =

X

tan θ

=

73.58

127.44.

tan 30 ◦

In summary, the vertical reactions are X 1 = X 2 = X 3 = 73.58 N, and the hori-

zontal reactions are −Y 1 = Y 2 = Y 3 = 127.44 N.

=