Equilibrium - General Physics I - Lecture Slides, Slides of Physics

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Last time: what if two or more

different forces act on lever arm?

Net Torque

• The net torque is the sum of all the torques produced by

all the forces

• Remember to account for the direction of the tendency for rotation
  • Counterclockwise torques are positive
  • Clockwise torques are negative

Where would the 500 N person have to

be relative to fulcrum for zero torque?

Example 2:

Given:

weights: w 1 = 500 N

w 2 = 800 N

lever arms: d 1 =4 m

Find:

d 2 =?

1. Draw all applicable forces and moment arms

2 2 2 (800 )(2 ) (500 )( ) 800 2 [ ] 500 [ ] 0 3. RHS LHS N m N d m N m d N m d m τ τ = − = − ⋅ ⋅ + ⋅ ⋅ = ⇒ = ∑ ∑

500 N

800 N

d 2 m 2 m

According to our understanding of torque there

would be no rotation and no motion!

N’

y

What does it say about acceleration and force? Thus, according to 2nd^ Newton’s law ΣF=0 and a=0! (^) N N Fi N N N ' 1300 ( 500 ) ' ( 800 ) 0 = ∑ = − + + −^ =

Axis of Rotation

• So far we have chosen obvious axis of rotation
• If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque - The location of the axis of rotation is completely arbitrary - Often the nature of the problem will suggest a

convenient location for the axis

• When solving a problem, you must specify an axis of

rotation

• Once you have chosen an axis, you must maintain that choice consistently throughout the problem

Center of Gravity (center of mass)

• The force of gravity acting on an object must be

considered

• In finding the torque produced by the force of gravity, all of

the weight of the object can be considered to be

concentrated at one point

Coordinates of the Center of Gravity

• The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object
• The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry.
• Often, the center of gravity of such an object is the geometric center of the object. i i i cg i i i cg m m y and y m m x x Σ Σ = Σ Σ =

Example:

Given:

masses: m 1 = 5.00 kg

m 2 = 2.00 kg

m 3 = 4.00 kg

lever arms: d 1 =0.500 m

d 2 =1.00 m

Find:

Center of gravity

Find center of gravity of the following system:

m kg kg m kg m kg m m m m mx m x m x m m x x i i i cg

1. 136

1 2 3 1 1 2 2 3 3 =

∑ ∑

Equilibrium, once again

• A zero net torque does not mean the absence of rotational

motion

• An object that rotates at uniform angular velocity can be under the influence of a zero net torque - This is analogous to the translational situation where a zero net force does not mean the object is not in motion

More on Free Body Diagrams

• Isolate the object to be analyzed
• Draw the free body diagram for that object - Include all the external

forces acting on the object

Example:

Given:

weights: w 1 = 100 N

length: l=10 m

angle: α=30°

Find:

f =?

n=?

P=?

1. Draw all applicable forces

P N

N P

PL

L

mg

86. 6

cos 30 sin 30 0 2 =

∑ = −^ =   τ

Torques: Forces:

n N F n mg f N F f P y x 100

∑ ∑ α

2. Choose axis of rotation at bottom corner (τ of f and n are 0!)

Note: f = μs n, so 0.^866

N

N

n f μ s mg

So far: net torque was zero.

What if it is not?

Torque and Angular Acceleration ( )

, so

tangentialacceleration :

,multiply by

a r α

Fr ma r

F ma r

t t t t t

α 2 Fr mr t =

torque τ dependent upon object and axis

of rotation. Called moment of

inertia I. Units: kg m^2

2 i i I ≡Σ mr τ = I α The angular acceleration is inversely proportional to the analogy of the mass in a rotating system

Newton’s Second Law for a Rotating

Object

• The angular acceleration is directly proportional to the net torque
• The angular acceleration is inversely proportional to the moment of inertia of the object
• There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object.
• The moment of inertia also depends upon the location of the axis of rotation

Σ τ = I α