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Equilibrium An object is in “Equilibrium” when: 1.There is no net force acting on the object 2.There is no net Torque (we’ll get to this later) In other.

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Presentation on theme: "Equilibrium An object is in “Equilibrium” when: 1.There is no net force acting on the object 2.There is no net Torque (we’ll get to this later) In other."— Presentation transcript:

1 Equilibrium An object is in “Equilibrium” when: 1.There is no net force acting on the object 2.There is no net Torque (we’ll get to this later) In other words, the object is NOT experiencing linear acceleration or rotational acceleration. We’ll get to this later

2 Static Equilibrium An object is in “Static Equilibrium” when it is NOT MOVING.

3 Dynamic Equilibrium An object is in “Dynamic Equilibrium” when it is MOVING with constant linear velocity and/or rotating with constant angular velocity.

4 Equilibrium Let’s focus on condition 1: net force = 0 The x components of force cancel The y components of force cancel

5 Condition 1: No net Force We have already looked at situations where the net force = zero. Determine the magnitude of the forces acting on each of the 2 kg masses at rest below. 60° 30°

6 Condition 1: No net Force ∑F x = 0and∑F y = 0 mg = 20 N N = 20 N ∑F y = 0 N - mg = 0 N = mg = 20 N

7 Condition 1: No net Force ∑F x = 0and∑F y = 0 ∑F y = 0 T1 + T2 - mg = 0 T1 = T2 = T T + T = mg 2T = 20 N T = 10 N T1 = 10 N 20 N T2 = 10 N

8 Condition 1: No net Force 60° mg =20 N f = 17.4 N N = 10 N F x - f = 0 f = F x = mgsin 60 f = 17.4 N N = mgcos 60 N = 10 N

9 Condition 1: No net Force ∑F x = 0and∑F y = 0 30° mg = 20 N T1 = 20 NT2 = 20 N ∑F y = 0 T1 y + T2 y - mg = 0 2T y = mg = 20 N T y = mg/2 = 10 N T2 = 20 N T2 x T2 y = 10 N 30° T y /T = sin 30 T = T y /sin 30 T = (10 N)/sin30 T = 20 N ∑F x = 0 T2x - T1x = 0 T1 x = T2 x Equal angles ==> T1 = T2 Note: unequal angles ==> T1 ≠ T2

10 Condition 1: No net Force ∑F x = 0and∑F y = 0 Note: The y-components cancel, so T1 y and T2 y both equal 10 N 30° mg = 20 N T1 = 20 NT2 = 20 N

11 Condition 1: No net Force ∑F x = 0and∑F y = 0 Note: The x-components cancel The y-components cancel 30° 20 N T1 = 40 N T2 = 35 N

12 30°60° Condition 1: No net Force A Harder Problem! a. Which string has the greater tension? b. What is the tension in each string?

13 a. Which string has the greater tension? ∑F x = 0 so T1 x = T2 x 30°60° T1 T2 T1 must be greater in order to have the same x-component as T2.

14 ∑F y = 0 T1 y + T2 y - mg = 0 T1sin60 + T2sin30 - mg = 0 T1sin60 + T2sin30 = 20 N Solve simultaneous equations! ∑F x = 0 T2x-T1x = 0 T1 x = T2 x T1cos60 = T2cos30 30°60° T1 T2 Note: unequal angles ==> T1 ≠ T2 What is the tension in each string?

15 Condition 2: net torque = 0 Torque that makes a wheel want to rotate clockwise is - Torque that makes a wheel want to rotate counterclockwise is + Negative TorquePositive Torque

16 Condition 2: No net Torque Weights are attached to 8 meter long levers at rest. Determine the unknown weights below 20 N ?? 20 N ?? 20 N ??

17 Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below 20 N ??

18 Condition 2: No net Torque F 1 = 20 N r1 = 4 mr2 = 4 m F 2 =?? ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(4)(sin90) = (20)(4)(sin90) F 2 = 20 N … same as F 1 Upward force from the fulcrum produces no torque (since r = 0)

19 Condition 2: No net Torque 20 N

20 Condition 2: No net Torque 20 N?? Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below

21 Condition 2: No net Torque ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(2)(sin90) = (20)(4)(sin90) F 2 = 40 N F 2 =?? F 1 = 20 N r1 = 4 mr2 = 2 m (force at the fulcrum is not shown)

22 Condition 2: No net Torque 20 N 40 N

23 Condition 2: No net Torque 20 N ?? Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below

24 Condition 2: No net Torque F 2 =?? F 1 = 20 N r1 = 3 mr2 = 2 m ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(2)(sin90) = (20)(3)(sin90) F 2 = 30 N (force at the fulcrum is not shown)

25 Condition 2: No net Torque 20 N 30 N

26 In this special case where - the pivot point is in the middle of the lever, - and ø 1 = ø 2 F 1 R 1 sinø 1 = F 2 R 2 sinø 2 F 1 R 1 = F 2 R 2 20 N 40 N 20 N 30 N (20)(4) = (20)(4) (20)(4) = (40)(2) (20)(3) = (30)(2)

27 More interesting problems (the pivot is not at the center of mass) Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg. Determine the unknown weight below. 20 N?? CM

28 Trick: gravity applies a torque “equivalent to” (the weight of the lever)(R cm ) T cm =(mg)(r cm ) = (100 N)(2 m) = 200 Nm More interesting problems (the pivot is not at the center of mass) 20 N ?? CM Weight of lever Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg.

29 Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg. Determine the unknown weight below. F1 = 20 N CM R1 = 6 m R2 = 2 m F2 = ?? Rcm = 2 m Fcm = 100 N ∑T’s = 0 T2 - T1 - Tcm = 0 T2 = T1 + Tcm F 2 r 2 sinø 2 = F 1 r 1 sinø 1 + F cm R cm sinø cm (F 2 )(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90) F 2 = 160 N (force at the fulcrum is not shown)


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