SITEMAP   School-college Physics Notes: Forces 5.4 Complex moments problem solving

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Forces 5 Turning forces: 5.4 Problem solving of more complex moment and equilibrium calculations

Doc Brown's Physics exam study revision notes

5.4 Problem solving of more complex moment and equilibrium calculations

These sorts of calculations are use in structural engineering building applications

Q1 If w1 is 12.5 N and 3.5 m from the pivot point, what weight w2 is required if placed at 2.5 m from the pivot to balance the beam?

Q2 A beam is placed evenly on a pivot point (fulcrum).

On one side a 10 N weight is placed 2 m from the pivot point and a 40 N weight a further 4 m from the pivot point.

How far from the pivot point must the centre of gravity of an 80 N weight be placed to perfectly balance the beam horizontally?

Q3 This calculation is the sort of thing civil engineers and architects have to consider in the construction of 'modern' buildings. A 5.0 m aluminium beam is suspended by a steel cable from a concrete beam and 3.0 m along rests on a steel pole.

Assuming gravitational field force is 9.8 N/kg, calculate T, the tension in newtons on the supporting steel cable. Q4 The diagram on the right shows downward forces (4000 N and 6000 N) acting on a concrete beam that is held in position by a wall at P - effectively acting as a pivot point.

Another 2nd supporting wall is to be erected at X.

Calculate the force F that this 2nd wall needs to be able to withstand to produce a structurally stable balanced equilibrium situation.

Keywords, phrases and learning objectives for turning forces

Be able to solve more complex problems involving moment and equilibrium calculations that may be needed in structural engineering building applications.

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Worked out ANSWERS to the moment calculations

These sorts of calculations are use in structural engineering building applications

Q1 If w1 is 12.5 N and 3.5 m from the pivot point, what weight w2 is required if placed at 2.5 m from the pivot to balance the beam?

anticlockwise moment = 12.5 x 3.5 = 43.75 Nm

clockwise moment = w2 x 2.5

To balance the moments must be equal so:

w2 x 2.5 = 43.75, therefore w2 = 43.75/2.5 = 17.5 N

Q2 A beam is placed evenly on a pivot point (fulcrum).

On one side a 10 N weight is placed 2 m from the pivot point and a 40 N weight a further 4 m from the pivot point.

How far from the pivot point must the centre of gravity of an 80 N weight be placed to perfectly balance the beam horizontally?

The principle of moments states that the sum of the clockwise moments must equal the sum of the anticlockwise moments to attain equilibrium. A moment (Nm) = F (N) x d (m)

The sum of the clockwise moments = (10 x 2) + (40 x {2 + 4}) = 20 + 240 = 260 Nm

To balance this the anticlockwise moment must = 240 Nm, 240 = 80 x d, d = 260/80 = 3.25

Therefore the 80 N weight must be placed on the left 3.25 m from the pivot point.

Q3 This calculation is the sort of thing civil engineers and architects have to consider in the construction of 'modern' buildings. A 5.0 m aluminium beam is suspended by a steel cable from a concrete beam and 3.0 m along rests on a steel pole.

Assuming gravitational field force is 9.8 N/kg, calculate T, the tension in newtons on the supporting steel cable.

The weight of the aluminium beam = 9.8 x 200 = 1960 N.

moment = force x perpendicular distance from turning point.

You consider the weight of the aluminium beam to act through its centre of mass 2.5 m from either end, but it is 0.5 m from the steel rod which is effectively the pivot point on which you base your moment calculations.

Its actually the same situation as a wheelbarrow described further down the page!

anticlockwise moment = beam weight x distance from steel pole to centre of mass of aluminium beam = 1960 x 0.5 = 980 Nm

clockwise moment = tension in the steel cable x distance from steel cable to steel pole = T x 3.0

at equilibrium, i.e. balanced, the clockwise moment = anticlockwise moment

therefore: T x 3.0 = 980, so T = 980/3 = 327 N (3 sf) Q4 The diagram on the right shows downward forces (4000 N and 6000 N) acting on a concrete beam that is held in position by a wall at P - effectively acting as a pivot point.

Another 2nd supporting wall is to be erected at X.

Calculate the force F that this 2nd wall needs to be able to withstand to produce a structurally stable balanced equilibrium situation.

Clockwise moment = 6000 x (40 + 20) = 360 000 N

Anticlockwise moments = (4000 x 32) + (40 x F)

360 000 = 128 000 + (40 x F)

232 000 = 40F

F = 232 000 / 40 = 5800 N

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