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Grade 9Mechanics

A uniform horizontal footbridge is 12 m long and weighs 4000 N. It rests on two supports X and Y. A man of weight 600 N is a distance of 4 m from support X. What is the upward force on the footbridge from support X?

Profile image of Shrey Thakkar
8 Years agoGrade 9
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2 Answers

Profile image of Rituraj Tiwari
5 Years ago
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
Profile image of ammar jamil
4 Years ago
All is good except the first declaration “Since the man sits from the support X then Y acts as the pivot.”.
reason: man standing anywhere on bridge will exert force on both supports X & Y and hence both will become pivots. But (and instead), we are using F from support X because that’s what question asks “what is the upward force on the footbridge from support X?