Presentation on theme: "Equilibrium An object is in “Equilibrium” when:"— Presentation transcript:
1Equilibrium An object is in “Equilibrium” when: There is no net force acting on the objectThere is no net Torque (we’ll get to this later)In other words, the object is NOT experiencinglinear acceleration or rotational acceleration.We’ll get to this later
2An object is in “Static Equilibrium” when it is NOT MOVING.
3Dynamic Equilibrium An object is in “Dynamic Equilibrium” when it is MOVING with constant linear velocityand/or rotating with constant angular velocity.
4Equilibrium Let’s focus on condition 1: net force = 0 The x components of force cancelThe y components of force cancel
5Condition 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 the2 kg masses at rest below.60°30°
6Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0mg = 20 NN = 20 N∑Fy = 0N - mg = 0N = mg = 20 N
7Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0T1 = 10 N20 NT2 = 10 N∑Fy = 0T1 + T2 - mg = 0T1 = T2 = TT + T = mg2T = 20 NT = 10 N
8Condition 1: No net Force N = mgcos 60N = 10 N60°mg =20 Nf = 17.4 NN = 10 NFx - f = 0f = Fx = mgsin 60f = 17.4 N
15Condition 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
16Condition 2: No net Torque Weights are attached to 8 meter long levers at rest.Determine the unknown weights below20 N??
17Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below20 N??
18Condition 2: No net Torque Upward force from the fulcrum produces no torque (since r = 0)∑T’s = 0T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1(F2)(4)(sin90) = (20)(4)(sin90)F2 = 20 N … same as F1r1 = 4 mr2 = 4 mF1 = 20 NF2 =??
26In this special case where - the pivot point is in the middle of the lever,and ø1 = ø2F1R1sinø1 = F2R2sinø2F1R1= F2R220 N40 N30 N(20)(4) = (20)(4)(20)(4) = (40)(2)(20)(3) = (30)(2)
27More 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
28More interesting problems (the pivot is not at the center of mass) Trick: gravity applies a torque “equivalent to” (the weight of the lever)(Rcm)Tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm20 N??CMWeight of leverMasses are attached to an 8 meter long lever at rest.The lever has a mass of 10 kg.
29Masses 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 NCMR1 = 6 mR2 = 2 mF2 = ??Rcm = 2 mFcm = 100 N∑T’s = 0T2 - T1 - Tcm = 0T2 = T1 + TcmF2r2sinø2 = F1r1sinø1 + FcmRcmsinøcm(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)F2 = 160 N(force at the fulcrum is not shown)