1. Static Equilibrium
(Serway )
Physics 1D03

2. Equilibrium of a Rigid Body
For a particle, “Equilibrium” means Fnet = 0, and then v = constant.
For extended 2-D or 3-D objects, this is not enough! F1
Here , so aCM = 0.
But the object is not in equilibrium:
It will spin (angular acceleration).
-F1
A “force couple”
Physics 1D03

3. Equilibrium of a Rigid Body
Net external forces and net external torque must be zero for a body in equilibrium.
1) a = 0 (no translational acceleration), so
(no net force)
2) a = 0 (no angular acceleration), so
(no net torque)
Physics 1D03

4. Example: B A f = 30° C Uniform beam AC: weight w1 = 100 N
length L = 3.00 m
Hinge at A, cable from B to C
w2 = 200 N A
f = 30° C
5/6 L
1/6 L
Find: tension in BC, force at A. w2
Steps:
1) Free-body diagram for the beam. The forces will be the weights w1 and w2, and the forces from the supports at A and C.
2) Net force = 0 (2 equations), net torque =0 (1 equation).
3) Solve for up to 3 unknowns.
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5. P T Forces on beam: θ 30° A C w1 w2 Py Px
Note the force P from the wall. The hinge at A prevents translational motion in any direction, so it may exert a force P in any direction.
The unknowns are: tension T, force |P|, and angle q ;
or, T, Px and Py .
Physics 1D03

6. ½ L w1 + 5/6 L w2 –LT sin30o = 0 30° T w2 w1 A C T sin30o
To find the tension, consider torques about A:
½ L w1 + 5/6 L w2 –LT sin30o = 0 +
30° T w2 w1 A Px Py C
5/6 L
½ L
T sin30o
so (after a little work),
T = w1 + 5/3 w2
= 433 N
To find the forces at A, consider the net force on the beam:
x components, Px - T cos30o = 0, so Px = T cos30o = 375 N
y components, Py + T sin30o - w1 - w2 = 0, so Py = 83 N
Physics 1D03

7. Question: θ 30° P T w2 w1 A C Py Px
How can you generate equations for Px and Py
which don’t involve the tension T?
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8. Answers:
1) torques about C: find Py (without using the value of tension!)
2) to find Px :
Take torques about B!
Py and T produce zero torque about B, so we get an equation in one unknown (Px ). B
30° T w2 w1 A Px Py C
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9. Equations of Equilibrium
In a 2-D problem, we can generate, at most, three independent equations from the requirements of static equilibrium of a single object. Possible combinations:
Two force equations,
one torque equation
2) One force equation (components along some axis)
two torque equations (torques about two different “pivots”).
3) Three torque equations (torques about 3 different “pivots”).
2) and 3) sometimes do not give 3 independent equations. You shouldn’t pick 3 collinear pivot points; and if you use two pivot points, taking force components along an axis perpendicular to the line through these two points will not give a third independent equation.
Physics 1D03

10. Static Friction Problems: “When does the ladder slip?”
Assume it is not slipping (so it is in equilibrium);
but is about to slip (so you can set fs = ms n, at each point that has to slip for motion to occur).
We may want to know:
For what angles q will the ladder slip? or
What coefficient of friction is required to prevent slipping? θ
Physics 1D03

11. Example: ladder
A uniform ladder of length L leans against a smooth (= frictionless) wall at angle q to the ground. What is the smallest coefficient of friction μs between ladder and ground which will prevent the ladder from slipping?
Consider the free body diagram. Note the force P from the wall is perpendicular (no friction from the wall). B θ P N w fs C
Plan: treat the weight w, length L, and angle q as known; use the equations of equilibrium to find the unknowns N and fs (and perhaps P). A
final answer:
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12. θ Or the minimum angle: Note that w, L cancelled.
Take torques about A: θ P N w fs B y x C
Then, net force is zero: A
Or the minimum angle:
Note that w, L cancelled.
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13. What if there is friction at both the floor and the wall?
What if someone is standing on the ladder? Does that change the risk of slipping?
What if there is friction at both the floor and the wall?
What if the floor is smooth and the wall is rough (has friction)? Can the ladder still stay in equilibrium for some range of angles?
Physics 1D03