A particular bridge, with a length L, consists of a horizontal slab with a weigh

Question

A particular bridge, with a length L, consists of a horizontal slab with a weight of 48.0 kN, with the weight uniformly distributed. The slab is held up by one support 20% of the way across the bridge, and a second support 80% of the way across the bridge. As shown in the picture, a truck with a weight of mg = 21.0 kN is driving from left to right across the bridge.





1) At the instant the truck’s weight is directly above the right support, what is the magnitude of the force the right support exerts on the horizontal slab?

______________kN

Explanation / Answer

ask if you have any doubt.

The Force equation will be FR + FL = 69.0 KN .

torque due to left support = FL(.3L) .

torque due to truck weight = 21.0KN (.3L) .

Torque due to right support  = FR(0.3L).

The torque balancing gives FR(.3L) + 21KN(.3L) = FL(.3L) gives FL - FR  = 21 KN .

The two equations are FR + FL = 69.0KN and FL - FR  = 21 KN . adding these two gives FR = (69 + 21)/2 = 45KN.

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