Transcript Analyzing Forces in Equilibrium

Forces in Static Equilibrium
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When an object is in static equilibrium
(velocity = 0 m/s) the net force acting on the
object is zero
To solve these problems, we must break up
the angled forces into their x and y
components
Let’s Try an Example Together!
T1
T2
30°
60°
20 kg
A 20 kg sign is
suspended
motionless from
the ceiling by two
ropes, T1 and T2.
Calculate the
unknown
tensional forces
(T1 and T2).
Step 1: Draw a simple sketch for the forces
acting on the hanging mass
T2
T1
30°
60°
W
Step 2: Measure the angle of each force
starting from the positive x axis and rotating
counterclockwise
T2
T1
+ x axis
W
Step 3: Write each force with their respective
angle
T2
T1
+ x axis
T1 @ 150°
T2 @ 60°
W @ 270°
W
Note: Weight
(Fgrav) is always
at a 270° angle
Step 4: Break each force down into its x and y
components using an organized table (use cos
for x and sin for y components)
Force
X
Y
T1 @ 150°
T1cos150°
T1sin150 °
T2 @ 60 °
T2cos60 °
T2sin60 °
W @ 270 ° (20)(9.8)cos270 ° (20)(9.8)sin270 °
Resultant
0
0
Step 5: Solve for any possible answers that you
can in the table
Force
X
Y
T1 @ 150°
T1cos150°
= -.866T1
T2cos60 °
= .5T2
(20)(9.8)cos270 °
=0
0
T1sin150 °
= .5T1
T2sin60 °
= .866T2
(20)(9.8)sin270 °
= -196
0
T2 @ 60 °
W @ 270 °
Resultant
Step 6: Solve for the unknown variables (T1 and
T2) by setting the sum of the x components
equal to zero and/or by setting the sum of the y
components equal to zero
We can do this because the system is in
equilibrium (constant velocity = 0 m/s) therefore
Fnet = 0
X components
-.866T1 + .5T2 + 0 = 0
.5T2 = .866T1
.5
.5
T2 = 1.732T1
Y components
Step 7: Now use the substitution method to take
your answer from the x component and use it to
help solve for T1 and T2
..5T1
+ .866T2 + (-196) = 0
.5T1 + .866(1.732T1) + (-196) = 0
.5T1 + 1.499T1 = 196
1.99T1 = 196
T1 = 98.5 N
Step 8: Now, if necessary, solve for the
other unknown force
T2 = 1.732T1
And since we know T1 = 98.5 N
T2 = 1.732(98.5)
T2 = 171 N