The figure shows a cross member of a bridge. The mass of the member is M = 125 k

Question

The figure shows a cross member of a bridge. The mass of the member is M = 125 kg and its center of mass is at its geometrical center. Two forces, (1) F (with x and y components) acting at A and (2) R (also with x and y components) acting at O, represent the contacts at the ends of the member. The angle = 41.7 and Ry = +2Mg.

(a) What is Rx (N)?

(b) What is Fx (N)?

(c) What is Fy (N)?

I know there is another question on the chegg that is very similar. The answer is wrong and the work shown is incorrect. Please do not copy and paste the wrong answer from that page. Thank you.

Explanation / Answer

sum torques about top end
- Ry L cos(41.7) + Rx L sin(41.7) + Mg L/2 cos(41.7)=0
-2*125*9.81*cos(41.7) + Rx *sin(41.7) + 125*9.81/2*cos(41.7)=0
Rx=2064 N

sum forces in the x
Rx + Fx = 0
Fx = - Rx = -2064 N

sum forces in the y
Fy + Ry - mg = 0
Fy + 2*125*9.81 - 125*9.81 = 0
Fy = -125*9.81=-1226 N

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