When forces acting on an object which is at rest are balanced, then we say that the object is in a state of static equilibrium.
The resultant of these forces equals zero. That is, the vector sum of the forces adds to zero.
Example #1
Suppose two dogs are struggling for the same shoe as shown in the diagram below. The left dog's force is shown by the yellow/black arrow while the right dog's force is shown by the teal arrow. Notice below that when the forces are added head-to-tail, the resultant force, shown in grey, acts straight up the y-axis.
Forceright dog + Forceleft dog = Resultant
Fr +
Fl =
R
To place the shoe into a state of static equilibrium, a third force (shown in red) would have to be added to the diagram. Then the forces exerted on the shoe would equal zero.
Forceright dog + Forceleft dog + "Equilibrium" vector = 0
Fr +
Fl + E = 0
Fr +
Fl =
-E
By comparing the vector equations presented above
Fr +
Fl = R
Fr +
Fl = -E we notice that the third force required for equilibrium to be established is a vector that has the same magnitude as the resultant, but points in 180º the opposite direction.
R = -E
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Each dog's force has been resolved into its x- and y-components.
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A vector diagram showing only the components of each dog's force.
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The addition of a third force would place the shoe into a state of static equilibrium.
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Example #2
Suppose you are asked to calculate the tensions in the three ropes (A, B, C) that are supporting the 5-kg mass shown below. Since the system is at rest, we will work the problem using the properties of static equilibrium.
Let's begin with a freebody diagram showing the forces acting on the knot. Since these forces belong to three separate ropes, all three tensions can be different.
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The next step will be to build an x|y chart showing the components of each force.
If θ were to equal 37º then |
Since we do not have any knowledge of any of the three tensions, we must now do a similar analysis using a freebody diagram of the 5-kg mass.
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The x|y chart for the 5-kg mass would look as follows:
Since we know that the 5-kg mass is in static equilibrium, we know that the sum of the forces in each column equals zero. We only need to write the equation for the y-column since there are no non-zero forces in the x-column.
y: C + (-mg) = 0
C - 5(9.8) = 0 C = 49 N
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Now, knowing the value for C, we can solve for the tensions in ropes A and B by setting up the following equations from each column in our x|y chart for the knot:
x:
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-A cos(37º) + B = 0
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y:
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A sin(37º) - C = 0
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x:
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-0.8A + B = 0
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y:
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0.6A - C = 0
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Since we know the value for C we will next solve for A and then for B.
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y:
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0.6A - 49 = 0 A = 49/0.6 A = 81.7 N
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x:
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-0.8(81.7) + B = 0 B = 0.8(81.7) B = 65.3 N
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Example #3
This procedure can also be applied to the following situation in which a mother is just at the instant of releasing her child in a swing.
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