A Classical Approach of Newtonian Mechanics

10 STATICS 10.6 Jointed rods

R truck S

l/3

l

which yields

(m g/2 + M g x/l)

S =

tan θ

=

(0.5 × 40 × 9.81 + 80 × 9.81 × 7/10)

tan 83.11◦

= 90.09 N.

The condition that zero net vertical force acts on the ladder yields

Hence,

R − m g − M g = 0.

R = (m + M) g = (40 + 80) × 9.81 = 1177.2 N.

Worked example 10.4: Truck crossing a bridge

Question: A truck of mass M = 5000 kg is crossing a uniform horizontal bridge

of mass m = 1000 kg and length l = 100 m. The bridge is supported at its two end-

points. What are the reactions at these supports when the truck is one third of

the way across the bridge?

M g m g (^) bridge
Answer: Let R and S be the reactions at the bridge supports. Here, R is the
reaction at the support closest to the truck. Setting the net vertical force acting
on the bridge to zero, we obtain
R + S − M g − m g = 0.
Setting the torque acting on the bridge about the left-most support to zero, we
get
M g l/3 + m g l/2 − S l = 0.