1 / 12

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!.

dieter-santana
Download Presentation

Forces in Static Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related