s6 static equilibrium notes

7/28/2019 S6 Static Equilibrium Notes

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Statics Introduction to statics

Static Equilibrium Equations

Examples:

Suspended beam

Hanging lamp

Ladder

Australian Institute of Sport


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Statics:

Statics is the study of systems that dont move.

Ladders, Stability of solid objects

Balanced objects

Buildings, Suspension bridges

Clifton Suspension BridgeChinese Golden Dragon Acrobats


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Statics:

Example: Evaluate the forces acting on a car parked on a hill.

xy

N

mg

f

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Car on Hill:

Use Newtons 2nd Law: FNET

= MACM

= 0

xy

N

mg

f

F 0

x: f - mgsin= 0

f = mg sin

y: N - mg cos = 0

N = mg cos


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Suspended beam:

Now consider a beam of mass Msuspended by two strings as

shown. We want to find the tension in each string:

L/2

L/4

Mx cm

T1 T2

Mg

F

0 First use

T1 + T2= Mg

This is no longer enough to

solve the problem!1 equation, 2 unknowns.

We need more information!!

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Suspended beam

We dohave more information:

We know the beam is not rotating!

NET=I= 0

The sum of all torques is zero.

This is true about any axiswe choose.

0

L/2

L/4

Mx cm

T1 T2

Mg


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Using Torque...

Choose the rotation axis to be along the zdirection (out of the

page) through the CM:

2 2 4 TL

The torque due to the stringon the right about this axis is:

1 12

TL

The torque due to the string onthe left about this axis is:

Gravity exerts no torque about the CM

L/2

L/4

Mx cm

T1 T2

Mg


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Using Torque...

Now

TL

TL

2 14 2

0

T T2 12

We already found that

T1 + T2= MgT Mg1

1

3

T Mg22

3

L/2L/4

Mx cm

T1 T2

Mg

0


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Approach to Statics:

0F 0

In general, we can use the two equations

to solve any statics problem.

When choosing axes about which to calculate torque, we can

sometimes be clever and make the problem easier....

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Juan and Bettina are carrying 80 kg block on a

massless 4 m board. Find the force in newtonsexerted by each to carry the block.

1. Juan 400 N, Bettina 400 N





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Juan and Bettina are carrying 80 kg block on amassless 4 m board. Find the force in newtons

exerted by each to carry the block.

Force: Torque:

mg

FJ FB


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Center of Mass & Statics

The center of mass is at the point where the system

balances!

Sum of all gravitational torques about an axis through the

center of mass is 0!

02211 dmdm

m1m2

+

d1 d2

CM

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02211 gdmgdm


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Mobile

A mobile hangs as shown below.

The rods are massless.

The mass of the ball at the bottom right is 1kg.

Find the unknown masses.

1 kg

1 m 3 m

1 m 2 m


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Solution

First figure out M1.

1 kg

1 m 3 m

1 m 2 m

M1

0)1()1)(3( 1 Mmkgm

kg3M1


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Solution

Now M2.

0)2()4)(1(

4

2

Mmkgm

kgT

M kg2 2

M2

1 kg

1 m 3 m

1 m 2 m

3 kg

T


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Solution

So finally

1 kg

1 m 3 m

1 m 2 m

3 kg

2 kg


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Statics

A 1 kgball is hung at the end of a rod 1 m long. The system balances

at a point on the rod 0.25 m from the end holding the mass.What is the mass of the rod?

1. 0.5 kg 2. 1 kg 3. 2 kg

1 kg

1 m

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Solution

The total torque about the pivot must be zero.

1 kg

The center of mass of the rod is at its center, 0.25 m to theright of the pivot.

X

CM of rod

Since this must balance the ball, which is the same distanceto the left of the pivot, the masses must be the same!

same distancemROD = 1 kg


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Example: Hanging Lamp

A lamp of mass Mhangs from the end of plank of mass m and

length L. One end of the plank is held to a wall by a hinge, and

the other end is supported by a massless string that makes an

angle with the plank. (The hinge supplies a force to hold the

end of the plank in place.)

What is the tension in the string?

What are the forces supplied by the

hinge on the plank?

hinge

M

m

L

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M

m

L/2

Fx

Fy

T

L/2

Mg

mg

y

x

x: T cos + Fx = 0

y: T sin + Fy - Mg - mg= 0

0 Then in the zdirection. choose the rotation axis tobe through the hinge so hinge forces

Fx and Fy will not be included

0LTsin-mg2

LLMg

F

0First


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Hanging Lamp...

So we have three equations and three unknowns:

Fx = -T cos

Fy = Mg + mg - T sin

M

m

L/2

Fx

Fy

T

L/2

Mg

mg

mg

2

LLMgLTsin

which we can solve to find:

sin

g2mM

T+

=

tan

g2mMFx

F mgy 1

2


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What are the two forces F1 and F2 exerted bya hand on a 5m long rod of mass m? Compare

them to a total force exerted by the hand.

5 m

0.1 m

m

F1

F2 mg

2.4m

Torque:

Force:


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Example: Ladder against smooth wall

A ladder (length L, mass m) leans against a smooth wall (no

friction between wall and ladder).

A static frictional force Fbetween the ladder and the floorkeeps it from slipping (S = 0.70)

The angle between the ladder and the floor is .

What angle is needed to prevent the ladder

from slipping?

L m

F


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Example: Ladder against smooth wall...

Consider all of the forces.

Gravity

Friction

Normal forces Nfand Nw by the floor and wallrespectively on the ladder.

UseF = 0

x: -Nw + F = 0

y: Nf - mg = 0

F = Nw

Nf = mg

m

F

mg

Nw

Nf

y

x


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Now using base of ladder as axis.

0

L/2

m

F

mg

Nw

Nf

)sin()90sin(2

0 WLNmgL

W

W

N

mg

Nmg

2

1)tan(

)sin()cos(2

10


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L/2

m

F

mg

Nw

Nf

WNmg

21)tan(

mgNF

FN

SfS

W

36

71.7.02

1

2

1

2

1)tan(

SSmg

mg


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Tipping a Box

A box is placed on a ramp in the configurations shown below. Friction

prevents it from sliding. The center of mass of the box is indicated bya blue dot in each case.

In which cases does the box tip over?

1. all 2. 2 & 3 3. 3 only 4. none

1 2 3

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Box

We have seen that the torque due to gravity acts as though all themass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass is

lowered, it will!

1 2 3

updown


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Box

1 23

We have seen that the torque due to gravity acts as though all themass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass is

lowered, it will!

s6 static equilibrium notes

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