5. A simple bridge of 36.5 m length has two concrete supports, one on each end. A car drives across the bridge and stops 10.2 m from the left end. The bridge weighs 2.56 x 105 N. The car weighs 5.25 x 104 N. If the bridge is in equilibrium, what force is each support providing

X = 2.882×10⁵ N,

Y =  1.43×10⁴ N

Explanation:

From the diagram attached,

Let the reaction from the two support be X and Y.

According to the condition of equilibrium

Sum of upward force = sum of downward force.

X+Y = 5.25×10⁴+2.5×10⁵

X+Y = 3.025×10⁵.............................. Equation 1

Also,

Sum of clockwise moment = sum of anticlockwise moment.

Take the moment about support A

(10.2×5.25×10⁴)+(18.25×2.56×10⁵) = 36.5×Y

52.075×10⁴ = 36.5Y

Y =52.075×10⁴/36.5

Y = 1.43×10⁴ N.

Substituting the value of Y into equation 1

X+ 1.43×10⁴ = 3.025×10⁵

X = 3.025×10⁵-1.43×10⁴

X = 2.882×10⁵ N.

Hence the force provided by each supports are

X = 2.882×10⁵ N, and Y =  1.43×10⁴ N

powerful computers now, not then

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )