A uniform horizontal footbridge is 12 m long and weighs 4000 N. It rests on two supports X and Y .If 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.
Answers
Answer:
The upward force = 2400 N
Explanation:
This is a question on moments.
Since the man sits from the support X then Y acts as the pivot.
Moments = Force × distance from the pivot
Distance of the man from the pivot X = 12 - 8 = 8m
The weight of the footbridge acts at the center of the bridge which is 6m from the pivot.
The upward force on the bridge is equal to the resultant force and acts at 12m from the pivot.
Let the Force be F.
Upward moments = downward moments
Downward moments = 8m × 600N + 6m × 4000 N
= 4800Nm + 24000Nm = 28800Nm
The upward moments = 12m × F = 12FNm
12F = 28800
F = 28800/12 = 2400 N
The upward force = 2400 N
Answer:
Explanation:
In last step it is divided by 12 sorry for that I have divided mistakenly by 2 ... answer is correct and concept is that one
