(r i F i ) F i = 0. C O = i=1

Transcription

1 Notes on Side #3 ThemomentaboutapointObyaforceF that acts at a point P is defined by M O (r P r O F, where r P r O is the vector pointing from point O to point P. If forces F, F, F 3,..., F N act on particles P,P,P 3,..., P n,thatarelocatedatpositionsr, r, r 3,..., r N, as measured from point O, then moment about a point O due to these forces is given by M O (r i F i. Note that is no force acts on, say particle P k,thenf k 0. A couple is a set of forces that add to zero, so that rx F i 0, if these forces act at positions r, r, r 3,..., r r,asmeasuredfrompointo, the couple of these forces about point O is defined as this need not be zero. C O rx (r i F i Notes on Side #4 Now the resultant force acting on a system of particle does not depend on the origin, but the resultant moment does depend on the choice of origin. In fact, if we use P r i P r O + O r i then we have M P ( P r i F k (( P r O + O r i F i ( O r i F i + P r O F i

2 or M P M O + P r O F. where r O/P is a vector pointing from P to O. When F 0, asinacouple,wehave C P C O, showing that a couple is independent of the choice of reference point (P or O. Notes on Side #6 Starting with M P M O + P r O F we note that M O M P P r O F. Given a non-zero vector B, we know that any vector (say A can always be written as a sum of two parts, one part parallel to B the other part perpendicular to B. To show this we start with A B + B where B k αb B B 0. This leads to or Then B A αb resulting in (A αb B 0 Note that we may write this as so we have B α B A B B. B A αb A (B BA (A BB B B A B A B + B B A B B. B B (B A B B B B A B. B B for any non-zero vector B. NotethatifB is an unit vector (i.e., b, then this reads A ³ b A b + ³ b A b.

3 If we apply this to M O F, wehave F MO F MO M O M O + M O F + F F F F comparing this to M O M P P r O F we see that by choosing P so that P r O F M O F F then we will have M O P r O F M P M O, which is then parallel to F. NotesonSides#7#8 Consider recall the expression A v P i d(a r P i dt Q j A ( A r P i + t nx j F (a i Av P i j A ( A r P i q j dq j dt rpi t + F. nx j ( A v P i j q j where A v P A ( A r P i A ( A v P i i j q j are the jth partial velocities of point P i, as measured by A. Suppose now that the system is a rigid body so that the particles P i are fixedintherigidbodyb suppose that P is some arbitrary point fixedinb.ifwethenchoosepasa reference point in B, we may use the fact that A v P i A v P + A ω B P r P i for each particle P i,where A ω B is the angular velocity of B as measured by A. Then we may write A v P i j (A v P i (A v P + A ω B P r Pi (A v P + (A ω B P r Pi 3

4 or where A v P i j (A v P + (A ω B P r P i + A ω B ( P r Pi A v P j + A ω B j P r P i + A ω B ( P r P i A v P j (A v P A ω B j (A ω B is defined as the jth partial angular velocity of B as measured by A, for j,, 3,...,n.Since P r P i are not functions of the q j s, we have so A v P i j ( P r P i 0 A v P j + A ω B j P r P i Putting this into the expression for Q j. i.e., we have or simply Q j F (a i (F (a Q j F Av P j + F Av P j + F Av P j + F (a i Av P i j ³A vj P + A ω B j P r P i i Av P j + X N (F (a i Aω B j P r P i A ω B j P r Pi F (a i A ω B j ( P r P i F (a i P r P i F (a i Aω B j Q j F Av P j + M P Aω B j 4

5 Notes on Side #0: A Slider Crank Mechanism Consider the Crank-Slider mechanism shown in the figure below. It consists of two uniform rods (a left rod a right rod connected to each other by a revolutejoint. Theleftendoftherodontheleft(thecrankisfreetorotate about a fixed revolute joint. The right end of the rod on the right is connected to a slider which is free to slide along a horizontal track. A simple crank-slider mechanism Therodontheleftisthecrank the particle on the right is the slider WeassumethereisstaticfrictionactingatsliderPthatithasmass m it is on the verge of slipping, we assume the links have masses m (left link m (right link that {ba, ba } is a SRT in A with ba pointing to the right ba pointing up. We assume that gravity points down in the negative ba direction. The crank makes an angle θ with ba the slider is a distance x from the origin (the revolute joint at point O at the bottom of the crank. The length of the crank is r the length of the right rod this rod makes an angle ϕ with ba. Suppose that there is a force F F ba acting on P (the slider there is a torque τ τ ba 3 5

6 acting about the pivot at point O (the end of the crank. We assume no friction in the axial so that axial forces do no work. Computing the generalized force associated with the coordinate q θ, wehave where Q F Av P + K Av P + W Av C + W Av C + M O Aω B W m gba W m gba are the weights of the left right rods, respectively, S Sba μ s N ba μ s mgba is the static friction acting on the slider P. But r P xba with x r cos(θ+l cos(ϕ, r sin(θ L sin(ϕ leading to sin(ϕ r L sin(θ, which says that Then ϕ cos(ϕ r L θ cos(θ or ϕ r L A v P ẋba ( r θ sin(θ L ϕ sin(ϕba θ cos(θ cos(ϕ, or Thus We also have M O τ ba 3 A v P r In addition, we have sin(θ cos(θ cos(ϕ sin(ϕ A v P (A v P θ A ω B θba 3 θba r θ sin(θ + ϕ ba. cos(ϕ r sin(θ + ϕ ba. cos(ϕ A ω B (A ω B θ ba 3. r C r cos(θ ba + r sin(θ ba 6

7 so that hence We also have A v C r C µx L cos(ϕ ba + L sin(ϕ ba A v C A d(r C dt r θ sin(θba + r θ cos(θba A v C A ( A v C θ r sin(θ ba + r cos(θ ba. A d(r C µẋ + dt L ϕ sin(ϕ ba + L ϕ cos(ϕ ba µ r θ sin(θ L ϕ sin(ϕ+ L ϕ sin(ϕ ba + L ϕ cos(ϕ ba µ r θ sin(θ L ϕ sin(ϕ ba + L ϕ cos(ϕ ba. But so we have A v C r θ sin(θ L r L ϕ r L θ cos(θ cos(ϕ cos(θ θ cos(ϕ sin(ϕ ba + L r cos(θ θ L cos(ϕ cos(ϕ ba which reduces to A v C r sin(θ cos(θ cos(ϕ sin(ϕ θba + r θ cos(θba so A v C A ( A v C θ r The generalized force is now sin(θ cos(θ cos(ϕ sin(ϕ ba + r cos(θ ba. Q F Av P + S Av P + W Av C + W Av C + M O Aω B r sin(θ + ϕ r sin(θ + ϕ ( F ba ba +(μ s mgba ba cos(ϕ cos(ϕ 7

8 which reduces to µ +( m gba r sin(θ ba + r cos(θ ba +( m gba r sin(θ cos(θ cos(ϕ sin(ϕ ba + r cos(θ ba +(τ ba 3 ba 3 Q τ + r(f μ smgsin(θ + ϕ cos(ϕ (m + m gr cos(θ If this crank-slider system is in static equilibrium, then we must have Q τ + r(f μ smgsin(θ + ϕ cos(ϕ (m + m gr cos(θ 0. Notes on Side # #: The Robotic Arm Consider the two-link robotic arm shown in the following figure. Each arm has mass m m, respectively center of mass positions at C C, respectively, as measured from the bottom of each arm. The Two-Link Robotic Arm The angles q θ q θ are measured from the dotted lines. Here we have two degrees of freedom, q θ q θ.wehave F F x ba + F y ba 8

9 acting at point P in the figure. The weights acting on each arm include these act at the points W m gba W m gba r C r C L cos(θ ba + L sin(θ ba µ L cos(θ + µ L cos(θ + θ ba + L sin(θ + L sin(θ + θ ba, respectively, so A v C L θ sin(θ ba + L θ cos(θ ba so that A v C µ L θ sin(θ L ( θ + θ sin(θ + θ ba µ + L θ cos(θ + L ( θ + θ cos(θ + θ ba, A v C L sin(θ ba + L cos(θ ba, A v C 0 A v C µ L sin(θ µ L sin(θ + θ ba + L cos(θ + L cos(θ + θ ba, A v C L sin(θ + θ ba + L cos(θ + θ ba. Now the torques τ τ are applied about the pivots at the origin at wherethetwoarmsmeettheforcef is applied at point P the moment M z is applied about point P. From the geometry of the problem, it is clear that the position of point P is r P (L cos(θ +L cos(θ + θ ba +(L sin(θ +L sin(θ + θ ba 9

10 so A v P ( L θ sin(θ L ( θ + θ sin(θ + θ ba +(L θ cos(θ +L ( θ + θ cos(θ + θ ba. Then or A v P A ( A v P θ A v P (L sin(θ +L sin(θ + θ ba +(L cos(θ +L cos(θ + θ ba. We also have A v P A ( A v P θ or We also have A ω B θ ba 3 A v P L sin(θ + θ ba + L cos(θ + θ ba. A ω B ( θ + θ ba 3 so that A ω B A ( A ω B θ ba 3, A ω B A ( A ω B θ ba 3, A ω B A ( A ω B θ 0 A ω B A ( A ω B θ ba 3, give the partial angular velocities of the rigid bodies (the two arms B B. The generalized forces now lead to Q F Av P + W Av C + W Av C + τ Aω B τ Aω B +τ Aω B + M P Aω B where τ acts at the bottom of the first arm τ acts at the top of the first arm. We also have Q F Av P + W Av C + W Av C + τ Aω B + τ Aω B + M P Aω B whichleadto Q F x (L sin(θ +L sin(θ + θ + F y (L cos(θ +L cos(θ + θ m g µ L cos(θ m g L cos(θ + L cos(θ + θ + τ + M z 0

11 Q F x L sin(θ + θ +F y L cos(θ + θ m g L cos(θ + θ +τ + M z At static equilibrium, we must have Q Q 0. Notethattheseareonly two equations (since there are two generalized coordinates as opposed to the 6 equations that would be needed using Newton s methods. Notes on Side #3: The Robotic Arm The student should do this problem again using the generalized speeds u ϕ θ u ϕ θ + θ given in the slide notice the difference.