Question

A uniform bridge span weighs 70 x 10^3 N and is 40 m long. An automobile weighing 15 x 10^3 N is parked with its center of gravity located 15 m from the right pier. What upward support force does the right pier provide?

Answer 1

44.375 × 10³ N

Step-by-step explanation:

Given:

Weight of the bridge, W = 70 × 10³ N

Length of the bridge, L = 40 m

Weight of the car, w = 15 × 10³ N

Location of the car from the right pier = 15 m

Now,

at equilibrium, the net torque on the left pier is zero

mathematically,

- W × (L/2) - w (40 - 15) + R × 40 = 0

here,

L/2 is the location of the center of gravity of the bridge from the left pier

R is the upward force by the right pier

Also,

- sign denotes the moment developed is anticlockwise

+ sign denotes the moment developed is clockwise

therefore,

- 70 × 10³ × (40/2) - 15 × 10³ × (40 - 15) + R × 40 = 0

or

- 1400 × 10³ - 375 × 10³ + 40 × R = 0

or

R = 44.375 × 10³ N