1 / 12

Equilibrium of Rigid Body

Equilibrium of Rigid Body. A rigid body is said to be in statical equlibrium when the resultant force acting on the body is zero and the resultant torsion at any point of the rigid body is zero Mathematically is Σ F = 0 and Σ τ = 0 If Σ F = 0 the body is in translational equilibrium

kimo
Download Presentation

Equilibrium of Rigid Body

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. Equilibrium of Rigid Body • A rigid body is said to be in staticalequlibrium when the resultant force acting on the body is zero and the resultant torsion at any point of the rigid body is zero • Mathematically is • Σ F = 0 and Σ τ = 0 • If Σ F = 0 the body is in translational equilibrium • If Σ τ = 0 the body is in rotational equlibrium

  2. Static Equilibrium can be classified : • 1. Stable Equilibrium. • Stable Equilibrium is characterized by the raise in object’s center of gravity whenever it is given a stimulus • When disturbance has gone, the object will return

  3. 2. Unstable Equilibrium • Unstable Equilibrium is characterized by the lowering in object’s center of gravity whenever it is given a stimulus • When disturbance has gone, the object does not return to initial position

  4. 3. Indifferent ( Netral ) Equilibrium • Indifferent Equilibrium is characterized by the in object’s center of gravity will not undergo any change in height whenever it is given a stimulus

  5. Stable conditional: 1. Σ F = 0 ⇒ Σ Fx = 0 Σ Fy = 0 Σ Fx = 0 N = T cosα Σ Fy = 0 w = f + T sin α 2. Σ τ = 0 T sin α T f F α N T cosα ½ L.w - L T sin α = 0 w

  6. Hitunglahtegangantalidangayapadaujungengsel . Beratbatang = 20N,berat papan=30N 37o SMA 1 KLATEN

  7. Batangberatnya=500 N, teganganpadatalihanyamampumenahan 3000 N , berapabeban max yang dapatditahanbatang. Tgθ=3/4 0,375 L 45o 0,625 L θ

  8. Batang yang disandarkan Syaratkesetimbangan: Σ F = 0 Σ Fx = 0 NA = fs Σ Fy = 0 NB =W 2. Σ τ = 0 darititik B L sin θ. NA - 1/2 Lcosθ.W=0 licin NA A F NB w θ kasar fs B

  9. Batang yang disandarkan fsA Syaratkesetimbangan: Σ F = 0 Σ Fx = 0 NA = fsB Σ Fy = 0 NB +fsA=w 2. Σ τ = 0darititik B L sin θ. NA - 1/2Lcosθ.W +Lcosθ. fsA=0 FA kasar NA A FB NB w kasar θ fsB B

  10. Berapakahgaya(F) minimum supaya bola terangkatkeatas,jikaberat bola 15 N. 2M F 0,4 M

  11. Sebuahtrukmassa 3 ton melintasisebuahjembatan yang panjangnya 50 m, jikatrukjaraknyadariujungawal 20 m. Berapakahgaya yang bekerjapadaujung-ujungjembatan,jikaberatjembatan 50.000N .

  12. Sebuahtanggamassanya 40 kg yang disandarkanpadadinding yang licindanterletakpadalantai yang kasardengankoefisiengesekan 0,5.Seorang yang massanya 60 kg menaikitangga. Berapakahtinggiorangitumenaikitanggasampaisaattepatakantergelincirtangganya. 10 m 8 m 6 m

More Related