topic 7 static equilibrium all - queen's u

APSC 111 Statics Page 6.2

Static Equilibrium

Chapter 12-1 to 12.5 Static Equilibrium

• What is Equilibrium

• The Conditions for Equilibrium

• Solving Statics Problems

• Stability and Balance


EquilibriumAn object is in equilibrium if it has “constant motion”. IE:

• If it is moving in a linear direction, it does not accelerate, it continues with constant velocity.

• If it is rotating, the angular velocity remains constant.

This can be expressed mathematically as momentum being constant:

constantp mv constantL I

Static EquilibriumAn object is in static equilibrium if it has no motion. So linear and angular velocities are constant and zero

0p 0L

0v 0


Conditions for Equilibrium

In order for an object to be in equilibrium:– The vector sum of all external forces acting on

the object must be zero.

– The vector sum of all torques acting on the object must be zero.

0 xF 0 yF 0 zF

0 x 0 y 0 z


Conditions for EquilibriumWe will normally have problems in which all of the forces act in a plane. (Say x and y). Then

In this case, all of the torques have to be along a z axis (They are always perpendicular to force and moment arm, which are in the x-y plane.) Therefore:

The torque can be about any convenient axis!

0 xF 0 yF

0 z


Problem Solving Strategies1. Isolate the object being analysed. Draw a clear FBD

showing all the external forces acting on the body. For systems with more than one object, you will need more than one FBD.

2. Establish a convenient coordinate system, then apply: and

3. Choose a convenient pivot point and apply:

Assign one sign to ccw torques, and the opposite sign to cwtorques.

0xF 0yF

0z

APSC 111 Statics .7

Example Problem• A uniform beam of mass m = 1.8 kg and length L is at

rest with its ends on two scales. A uniform block of mass M = 2.7 kg is at rest on the beam with its center a distance L/4 from the beam’s left end. What do the scales read?


Step 1: Draw the free body diagram. In this case there are two normal forces (one at each end of the beam, plus the mass of thebeam and the mass of the block.

Take +y up. +x to the right, and let it pivot about the left end of the beam. Take torque cw to be positive.


Note: The normal forces at the left and right ends are what we are seeking. They will be the scale readings.

0 0y l rF F F Mg mg

4 20 (0) 0z l rF F Mg mg

4 2rF Mg mg 1 14 2rF Mg mg

1

Insert into 1 l rF Mg mg F 1 14 2( )lF Mg mg Mg mg

3 14 2lF Mg mg


Plug in numbers: M=2.7 kg, m=1.8 kg

1 1 1 14 2 4 22.7(9.8) 1.8(9.8) 15.4 NrF Mg mg

3 31 14 2 4 22.7(9.8) 1.8(9.8) 28.7 NlF Mg mg

Scale will read in kg

2.9 kglM 1.6 kgrM


Example ProblemA uniform ladder of length land weight 50 N rests against a smooth vertical wall. If the coefficient of static friction between the ladder and the floor is 0.40, find the minimum angle at which the ladder does not slip.


fF

wN

fNmg

0 0y fF N mg

0 0x w fF N F

0 sin cos 02z wLN L mg

Take the pivot point to be where the ladder touches the floor. Take cw rotations as positive.

Take +x right, +y up

1

2

3


f w sN mg N mg w f w s fN F N N

sin cos 2wLN L mg

Then f s fF N

1

2

3

12 cotwN mg

12 cot smg mg

1min cot 2 51.3s

Note that the result is independent of length or mass… it only depends on coeff of friction.


Example Problem

A firefighter’s ladder of mass m = 20 kg and length L = 12 m leans against a frictionless wall. It makes an angle of 60° with the pavement on which the lower end rests. The ladder’s center of mass is at the midpoint. A firefighter having a mass M = 80 kg climbs the ladder. At what point will the ladder slip if the coefficient of static friction is 0.4?


Take xL to be the position of the climber as a fraction of L

with (0,1)xL x

fF

wN

fN

mgMg


( ) ( )f w sN m M g N m M g

w f w s fN F N N Then, as before f s fF N

1

2

0 0y fF N mg Mg

0 0x w fF N F

0 sin cos cos 02z wLN L mg MgxL

1

2

3

sin cos cos2wLN L mg MgxL 3

sin ( ) cos2wmN Mx g ( ) cot

2wmN Mx g


Hence

( ) cot ( )2 sm Mx g m M g

( ) ( ) tan2 sm Mx m M

Note: if M=0, we recover the answer from previous problem.

Solve for x at the point of slipping.

( ) tan 2s m M mxM

0.4(20 80) tan 60 20 2 .7780

x

When the fireman gets 77% of the way up the ladder, it will slip !!


Example Problem• A uniform horizontal beam of weight 200 N

and length 8.00 m is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0 with the beam. If a 600 N person stands 2.00 m from the wall, find:

a. The tension in the cable, andb. The force exerted by the wall on the beam.


2m

8m


mgMg

T

yN

xN


0 sin 0y yF N T Mg mg 0 cos 0 cosx x xF N T N T

0 4 2 8 sin 0z mg Mg T Three equations, 3 unknowns.

sin2 4

1 200 600 1 313 N2 4 sin 2 4 sin

mg MgT

mg MgT

1

2

3


600 N 200 N 313sin 53 550 NyN

cos 313cos53 188 NxN T

Then get the forces on the beam:

1

2

2 2 2 2188 550 =581 NTotal x yF N N


Example Problem• A safe of mass M = 430 kg hangs by a rope

from a boom with dimensions a = 1.9 m and b = 2.5 m. The boom consists of a hinged beam and a horizontal cable that connects the beam to the wall. The uniform beam has a mass m = 85 kg. The mass of the cable and rope are negligible.

a. What is the tension Tc in the cable?b. Find the magnitude F of the net force on the

beam from the hinge.


0 0y v rF F T mg

0 0x h cF F T

20 cos sin cos 0Lz C rmg LT LT

Three equations, 4 unknowns. Get Tr from free body diagram of safe.

0 0y rF T Mg Mg

rT

rT Mg


2sin cos cosLC rLT mg LT

Now get the forces on the beam from the hinge:

12sin cos cosCT mg Mg

12( ) cotCT mg Mg

12

2.5( ) [.5(85) 430](9.8) 6093 N1.9C

bT mg Mga

0 0 6093 Nx h c hF F T F 0 0y v rF F T mg

(85 430)9.8 5047 Nv rF T mg Mg mg


2 2 2 26093 5047 =7912 NTotal h vF F F

TotalF

topic 7 static equilibrium all - queen's u

Documents