Static EquilibriumChapter 12

Very often we are interested in the question, will an object stay where it is? or will it move?

This is basically the question of is the object in equilibrium (though an object in equilibrium can continue moving or rotating at a constant rate).

For an object to be in equilibrium the sum of the forces an the sum of the torques on it must be zero.

Equilibrium

X~F = 0

X~⌧ = 0

Example 12-3A board of mass M=2.0kg serves as a seesaw for 2 children. Child A has a mass of 30 kg and sits 2.5m from the pivot point P (his center of gravity is 2.5m from the pivot). At what distance x from the pivot must child B of mass 25kg place herself to balance the seesaw?

A B2.5m x

mAg mBg

FN

Mg

knownmA = 30kgmB = 25kgM = 2.0kgxA=2.5m

unknownxB=? physics

X~F = 0

X~⌧ = 0

FN �mAg �mBg �Mg = 0 mAgxA �mBgxB = 0

xB =mA

mBxA =

30kg

25kg2.5m = 3.0m

Example 12-6A 5.0m ladder leans against a smooth wall at a point 4.0m above a cement floor. The ladder is uniform and has a mass m=12.0kg. Assuming the wall is frictionless (but the floor is not), determine the forces expected on the ladder by the floor and the by the wall.

5m

3m

4mmg

Fcx

Fcy

Fw

knownm = 12.0kg

X~F = 0

X~⌧ = 0

x: Fcx � Fw = 0 => Fcx = Fw

y: Fcy �mg = 0 => Fcy = mg

torque: choose axis at bottom of ladder(4.0m)Fw � (1.5m)mg = 0 Rsinθ from triangles

Fw =1.5m

4.0mmg =

1.5m

4.0m(12.0kg)(9.8m/s2) = 44N

Fcx = Fw = 44N

Fcy = mg = 118N

Fc =q

F 2cx + F 2

cy

=p

(44N)2 + (118N)2 = 126N

Fast and Furious (2001)

Is this possible?

mg

FN

Ffr

wheelbase = 3mwheel diameter = 0.33m

Car is a 1970 Dodge Chargermass = 1650kg

This is not a case of equilibrium, but the only motion is in the x-direction. So ⌃Fy = 0 ⌃⌧ = 0

FN = mg

θ

mg(1.5m) sin ✓ = Ffr(0.165m)

Ffr = 9mg sin ✓

if θ=30 then

Ffr = 4.5mg = 72, 765N

ax = 4.5g = 44.1m/s2

Homework

Chapter 12 - 32, 42, 69, 70