Can someone help? The figure shows a cross-member of a bridge lying in the xy pl

Question

Can someone help?

The figure shows a cross-member of a bridge lying in the xy plane of the page. The x- direction is horizontal and the y-direction is vertical. Gravity pulls down on the cross- member in the -y-direction. The mass of the member is M = 142 kg and its center of mass is at its geometrical center. Two forces, (1) F (with x and y components) acting at A and R (also with a- and y components) acting at O, represent the contacts at the ends of the member. The angle theta = 41.7degree and R_y = +2Mg. (a) what is R_x? N(plusminus 20 N) (b) What is F_x? N(plusminus 20 N) (c) What is F_y? N(plusminus 20 N).

Explanation / Answer

Here

Rx = -Fx

Fx is the horizontal component of the force of member of bridge

So        Fx= mgcos

Here given m=103kg, = 41.7o

Fx = (103)x (9.81) cos (41.7o)

            = 1010.43 x0.7466 = 754.4N

Fx = 754.4N

F= 754.4N

Fy = mg sin

            = 103 x 9.81 x sin 41.7o

            = 672.2N

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