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Forces in Static Equilibrium

Forces in Static Equilibrium. 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!.

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Forces in Static Equilibrium

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  1. Forces in Static Equilibrium

  2. 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

  3. Let’s Try an Example Together! A 20 kg sign is suspended motionless from the ceiling by two ropes, T1 and T2. Calculate the unknown tensional forces (T1 and T2). T1 T2 30° 60° 20 kg

  4. Step 1: Draw a simple sketch for the forces acting on the hanging mass T2 T1 30° 60° W

  5. Step 2: Measure the angle of each force starting from the positive x axis and rotating counterclockwise T2 T1 + x axis W

  6. Step 3: Write each force with their respective angle T1 @ 150° T2 @ 60° W @ 270° T2 T1 + x axis W Note: Weight (Fgrav) is always at a 270° angle

  7. 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)

  8. Step 5: Solve for any possible answers that you can in the table

  9. 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

  10. X components -.866T1 +.5T2 + 0 = 0 .5T2 = .866T1 .5 .5 T2 = 1.732T1

  11. 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

  12. 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

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