A uniform horizontal footbridge is 12 m long and weighs 4000 N. It rests on two supports X and Y as shown. A man of weight 600 N is at a distance of 4 m from support X. What is the upward force on the footbridge from support X?
Correct answer: C. 2400 N
- A. 2200 N
- B. 2300 N
- C. 2400 N
- D. 2600 N
Explanation
As the total length of the bridge is 12m and a man of 600N weight is standing at a distance of 4 m from X. The torque acting at a 4m distance is τ1 =12F Nm. The distance of Y from the center of the bridge is 6m. The weight of the bridge acting from its center is 4000N.So, the torque acting at a 6m distance is 6(4000).=24000Nm Since the weight of the bridge is acting downward and the torque is clockwise and negative so τ2 = -24000NmThe distance of man from Y is 8m.So the torque acting at 8m distance is 8(600)= 4800NmSince the force due to man is acting downward and torque is clockwise and negative so τ3 =-4800 NmApplying 2nd condition of equilibrium;∑τ = 0τ1 + τ2 + τ3 =012F -24000 - 4800 =012F = 24000 + 480012 F = 28800F = 28800/12F = 2400N
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