1. Mechanics Tutorial 5 Bridges Monday, 25 February 2019
Leeds City College

2. Bridges
The simplest bridge is a uniform plank that crosses a space between two supports.
We use the principle of moments to work out the forces that act on each support. mg
Monday, 25 February 2019
Leeds City College

3. Moments are involved… 2d mg Fucking Hell! Monday, 25 February 2019
Leeds City College

4. Moments on a bridge A B mg
There is a clockwise moment about A, and an anti-clockwise moment about B
Clockwise Moment = mg × d
Anticlockwise moment = mg × d
Since force = moment ÷ distance, the force on each pivot, A and B, is given by:
F = mg/2 mg A B 2d
Monday, 25 February 2019
Leeds City College

5. Putting a load on a bridgemg A B 2d m1
m1g x
The centre of mass is in the same place, but we now have an extra load of m1g on the bridge.
Force on A = moment about B ÷ distance between A and B
Taking moments about B:
Force on A = Moment about B ÷ 2d = (mg × d) + (m1g × x) 2d
Moment = (mg × d) + (m1g × x)
Force on B = moment about A ÷ 2d = (mg × d) + (m1g × [2d – x]) 2d
Taking moments about A:
Moment = (mg × d) + (m1g × [2d – x])
Monday, 25 February 2019
Leeds City College

6. Ladder S A ladder can be treated using moments
The ladder is stable because all four faces form a closed rectangle
F is the friction to hold the ladder against the wall.
Length, l
Normal force, N
W = mg q F A
Monday, 25 February 2019
Leeds City College

7. Ladder B S We can resolve vertically and horizontally: N = W and S = F
Anti-clockwise moment about A
= S × l sin q
Length, l
Clockwise moment about A
= W × l/2 cos q
Normal force, N
Principle of moments:
S × l sin q = W × (l/2) cos q
W = mg
Therefore: q F A
Monday, 25 February 2019
Leeds City College