Unformatted text preview: systems that don’t
move. Ladders, signposts, balanced beams, buildings, bridges... r
r
The key equations are familiar to us: ∑ F = ma r
r
∑τ = Iα r
r
If an object doesn’t move: a = 0 τ = 0
If r
∑ F = 0 The net force on the object is zero
r
The
∑ τ = 0 The net torque on the object is zero
(for any axis) Physics 211 Lecture 17, Slide 4 Statics:
Statics:
Example:
What are all of the forces acting on a car parked on a hill?
What y x N
f θ mg
Physics 211 Lecture 17, Slide 5
Physics 211 Lecture 17, Slide Car on Hill:
Car on Hill:
• Use Newton’s 2nd Law: FNET = MACM = 0
Use
NET
CM
• Resolve this into x and y components:
Resolve x: f  mg sin θ = 0
f = mg sin θ y: N  mg cos θ = 0 y x N
f N = mg cos θ θ mg
Physics 211 Lecture 17, Slide 6
Physics 211 Lecture 17, Slide Torque due to Gravity
Torque due to Gravity
Magnitude:
R⊥ r
RCM τ = Rcm Mg sin(θ )
= MgR⊥ Lever arm Physics 211 Lecture 17, Slide 7 Example:
Example:
Now consider a plank of mass M suspended by two strings as
Now
shown. We want to find the tension in each string:
shown. T1 T2
M x cm
L/4 L/2 Mg y
x
Physics 211 Lecture 17, Slide 8 Balance forces:
Balance forces:
r
∑F = 0
T1 + T2 = Mg
T1 T2
M x cm
L/4 L/2 Mg y
x
Physics 211 Lecture 17, Slide 9 Balance torques
r
∑τ = 0 Choose the rotation axis to be out of the
page through the CM:
page The torque due to the string
on the right about this axis is: L
τ 2 = T2
4
The torque due to the string on
the left about this axis is: L
τ1 = −T1
2 T1 T2
M x cm
L/4 L/2 Mg y Gravity exerts no
torque about the CM x Physics 211 Lecture 17, Slide 10 Finish the problem
The sum of all torques must be 0: L
L
−T1 + T2 = 0
2
4 τ1 + τ 2 = 0
T 2 = 2T1 T1 M x cm We already found that
T1 + T2 = Mg
T1 = T2...
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 Fall '10
 Tuceryan
 Physics, Force

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