Structures & Materials

VCE Physics.comStructures -

• Torque• Centre of mass• Equilibrium• Stability• Tension structures• Arches• Cantilevers• Ties

Structures

2

keynote:/Users/T02556857/Documents/Physics/Structures%20%26%20Materials/Keynotes%20%26%20podcasts/Structures%20%26%20equilibrium/Structures%20%26%20equilibrium.key?id=BGSlide-12





Torque

• Torque is the “turning effect”• “Moments” are the product of force and radius. These can cause

rotation.• Torque is a vector cross product - it will be perpendicular to F and r.

3

! =Fr sin"

10 N

2.0 m

! =2.0m"10N"sin60°=17Nm

60°

10 N x sin 60° = 8.7N

! =20Nm


Rotational equilibrium

• For the see-saw to be in balance the overall moments must add to zero.

4

10 kg

100 N

1.0 m

100 Nm (clockwise)

0.5 m

20 kg

200 N

100 Nm (anti-clockwise)

The centre of mass is over the fulcrum


Centre of mass

5

• The centre of mass is the point at which an object will balance.• This is rotational equilibrium - the sum of the torque around that point is

zero.• A force applied through the centre of mass will not cause a rotation.

30 kg50 kg

2.0 m

4.0 m

x =

m1x1+m2x2 + .....m1+m2 + .....

x =

(50kg!2.0m)+(30kg!4.0m)50kg+30kg x =2.75m

x =

220kgm80kg

x


Equilibrium

• For a structure to remain stationary:• The forces must be in equilibrium !F = 0 • The torques must be in equilibrium !" = 0

6

100 kg

1000 N

6.0 m4.0 m

200 N

Beam mass = 20kgR1 R2

3.0 m


Equilibrium calculations

7

0=(1000N!4m)+(200N!3m)"(R2!6m)

R2!6m =4000Nm+600Nm

R2 =

4600Nm6m

R2 =770N

Find R2: Torque = 0

0=(R1+770N )!(1000N+200N )

R1=1200N !770N

R1=430N

Find R1: Forces = 0


Stability

• As long as the centre of mass is above part of the base, the structure will not topple.

• This is because the torque acts to return the structure to its original position.

8

Rotational equilibrium

Torque will maintain stability, rotating back towards centre

of mass.

Torque will cause the box to continue to topple,

away from centre of mass.


Arches

• A beam will bend if a load is applied between the supports.• This puts the top in compression & bottom under tension.• It is liable to crack under tension.

9

Tension

Compression


• An arch transmits loads horizontally to the ends. Stone blocks are under compression.

• The arch must be supported horizontally eg by walls or buttresses.

Arches

10

Load

Reactions Reactions


Tension structures

• The vertical & horizontal forces must all be in balance.

11

100 N

TT

mg

35°35°

2T sin! =mg

2T sin35°=100N

T =

100N2sin35°

T =87N

50N50N

71N71N

As the vertical angle ! 0°, the tension increases.

87N87N


Cantilevers

• The sum of the torque around the entry point must be zero.

12

100 kg

1000 N200 N

Beam mass = 20kg

2.5 m

1.0 m

0.5 m

4.0 m

F

R

0=(200N!1.0m)+(1000N!2.5m)"(F !0.5m)

F =

200Nm +2500Nm0.5m

F =5400N

0=R !5400N !200N !1000N

R =6600N

Take torques around here!


1000 N

Ties

13

100 kg

1000 N

200 N

Beam mass = 20kg

2.0 m

4.0 m

0=(200N!2m)+(1000N!4.0m)"(T sin30°!4.0m)

T = 400Nm+4000Nm

0.5!4.0m

T =2200N

Take torques around here!

30°

T1100N

1900N

Force in beam is

Fh =1900N Fv =100N

Components of tension:

Tv =T sin30°=1100N

Th =T cos30°=1900N

Structures & Materials

Documents

Transcript of Structures & Materials