Physics 207: Lecture 26. Announcements. Make-up labs are this week Final hwk assigned this week, final quiz next week.

Transcription

1 Torque due to gravit Rotation Recap Phsics 07: ecture 6 Announceents Make-up labs are this week Final hwk assigned this week, final quiz net week Toda s Agenda Statics Car on a Hill Static Equilibriu Equations Eaples: Suspended bea Hanging lap adder 1 Torque due to Gravit As we now know τ = Iα where τ i = r i X F i i i Take the rotation ais to be along the z direction (as usual) and recall that τ i i So: τ i = τ Z,i = r X,i F Y,i -F X,i r Y,i = i (- i g) - 0 = g = gm i τ = Mg c i i c (also obtain via closest approach ethod) z-ais r 4 4 F 4 r 1 F 1 1 Where: M = i i 3 r 3 r F 3 F Page 1

2 Torque due to Gravit... So for the purpose of figuring out the torque due to gravit, ou can treat an object as though all of its ass were located at the center of ass. τ = NET Mg c M = i i CM M r c c Mg 3 New Section - Statics: As the nae iplies, statics is the stud of sstes that don t ove. adders, sign-posts, balanced beas, buildings, bridges, etc... Eaple: What are all of the forces acting on a car parked on a hill? N f g 4 Page

3 Car on Hill: Use Newton s nd aw: F NET = MA CM = 0 Resolve this into and coponents: F = 0 : f - g sin = 0 f = g sin : N - g cos = 0 N = g cos N f g 5 Using Torque: Now consider a plank of ass M suspended b two strings as shown. We want to find the tension in each string: First use F = 0 T 1 T T 1 + T = Mg c M This is no longer enough to solve the proble! 1 equation, unknowns. We need ore inforation!! / Mg /4 6 Page 3

4 Using Torque... We do have ore inforation: We know the plank is not rotating! τ NET = Iα = 0 τ = 0 T 1 T c M The su of all torques is zero! This is true about an ais we choose! / Mg /4 7 Using Torque... Choose the rotation ais to be along the z direction (out of the page) through the CM: The torque due to the string T 1 T on the right about this ais is: c τ = T 4 / /4 The torque due to the string on the left about this ais is: τ 1 = T 1 Mg Gravit eerts no torque about the CM M 8 Page 4

5 Using Torque... Since the su of all torques ust be 0: T T 1 =0 4 T = T 1 We alread found that T 1 + T = Mg T T 1 T 1 T 1 = Mg 3 = 3 Mg / c Mg /4 M 9 Approach to Statics: In general, we can use the two equations F = 0 τ = 0 to solve an statics proble. When choosing aes about which to calculate torque, we can be clever and ake the proble eas Page 5

6 Coent: Center of Mass & Statics The center of ass is at the point where the sste balances! Su of all gravitational torques about an ais through the CM = 0! d 1 1 d = 0 d d 1 = 1 d 1 d + 1 CM 11 ecture 6, Act 1 Statics A (static) obile hangs as shown below. The rods are assless and have lengths as indicated. The ass of the ball at the botto right is 1kg. What is the total ass of the obile? (a) 5 kg (b) 6 kg (c) 7 kg 1 1 kg Page 6

7 ecture 6, Act 1 Solution First figure out M 1. ( 3 )( 1 kg ) = ( 1 ) M 1 = 3 kg M 1 M 1 M 1 1 kg ecture 6, Act 1 Solution First figure out M 1. ( 3 )( 1 kg ) = ( 1 ) M 1 = 3 kg M 1 So the botto part has a total ass of 4 kg. kg 1 3 kg 1 kg Page 7

8 ecture 6, Act 1 Solution Now figure out M. ( 1 )( 4 kg ) = ( ) M = kg M kg 1 4 kg 15 ecture 6, Act 1 Solution So the total ass of the obile is: 1 kg + 3 kg + kg = 6 kg M 1 3 kg 1 kg Page 8

9 ecture 6, Act Statics A 1 kg ball is hung at the end of a rod 1 long. The sste balances at a point on the rod 0.5 fro the end holding the ass. What is the ass of the rod? (a) 0.5 kg (b) 1 kg (c) kg 1 1 kg 17 ecture 6, Act Solution A The total torque about the pivot ust be zero. The center of ass of the rod is at its center, 0.5 to the right of the pivot. Since this ust balance the ball, which is the sae distance to the left of the pivot, the asses ust be the sae! 1 kg sae distance X CM of rod ROD = 1 kg 18 Page 9

10 ecture 6, Act Solution B Since the sste is not rotating, the -coordinate of the CM of the sste ust be the sae as the pivot. The center of ass of the rod is at its center, 0.5 to the right of the pivot. Since the CM of the ball is 0.5 to the left of the pivot, the ass of the rod ust be 1 kg to ake CM = 0. 1 kg X CM of rod ROD = 1 kg 19 Eaple: Hanging ap A lap of ass M hangs fro the end of plank of ass and length. One end of the plank is held to a wall b a hinge, and the other end is supported b a assless string that akes an angle with the plank. (The hinge supplies a force to hold the end of the plank in place.) What is the tension in the string? What are the forces supplied b the hinge on the plank? M hinge 0 Page 10

11 Hanging ap... First use the fact that F = 0 in both and directions: : T cos + F = 0 : T sin + F - Mg - g = 0 Now use τ= 0 in the z direction. If we choose the rotation ais to be through the hinge then the hinge forces F and F will not enter into the torque equation: Mg + g - Tsin = 0 T / M g Mg / F F 1 Hanging ap... So we have three equations and three unknowns: T cos + F = 0 T sin + F - Mg - g = 0 Mg + g - Tsin = 0 which we can solve to find: ( M + ) g T = sin ( M + ) g F = tan F = 1 g T / M g Mg / F F Page 11

12 Eaple: adder against sooth wall Bill (ass M) is clibing a ladder (length, ass ) that leans against a sooth wall (no friction between wall and ladder). A frictional force F between the ladder and the floor keeps it fro slipping. The angle between the ladder and the wall is. What is the agnitude of F as a function of Bill s distance up the ladder? Bill F 3 Eaple: adder against sooth wall... Consider all of the forces acting. In addition to gravit and friction, there will be noral forces N f and N w b the floor and wall respectivel on the ladder. Again use the fact that F NET = 0 in both and directions: / N w : N w = F : N f = Mg + g g d Mg F N f 4 Page 1

13 Eaple: adder against sooth wall... Since we are not interested in N w, calculate torques about an ais through the top end of the ladder, in the z direction. sin g + ( d ) sin Mg + sin α F - sin N 0 cos f = ais / N w Substituting in N f = Mg + g and solving for F: d F = Mg tan + M α F d N f g α Mg 5 Eaple: adder against sooth wall... We have just calculated that d F = Mg tan + For a given coefficient of static friction µ s, the aiu force of friction F that can be provided is µ s N f = µ s g(m + ). M The ladder will slip if F eceeds this value. Morals: Brace the botto of ladders! Don t ake too big! d F 6 Page 13

14 Hinged Beas: Consider a structure ade fro two beas, attached to each other and the wall with hinges: = hinges 7 Hinged Beas... What we want to find: (A,A ) and (B,B ). A What we know: An forces present at C will act in pairs, and will therefore cancel if we consider the entire structure. A C g B B 1 g 8 Page 14

15 Hinged Beas... First use F NET = a in and directions: A + B = 0 A = -B (a) A A + B = ( 1 + )g (b) A That s two equations, but we have four unknowns... g B B 1 g 9 Hinged Beas... Now use soe torque relationships. First, consider the torque on the whole structure about an ais though the hinge at B. 1 g + g tan A = 0 g ( 1 + ) = tan A ( + ) g 1 = tan A (c) (No torques fro forces at C) C g 1 g A tan B 30 Page 15

16 Hinged Beas... If we knew soething about A or B we would be done! Do the siplest thing we can think of! Consider the torque on the botto bea about an ais through C: B g 1 = 0 g B = 1 (d) C B 1 g 31 Hinged Beas... So we have the following equations: (a) A = -B A (b) A + B = ( 1 + )g A (c) ( + ) g 1 = tan A B (d) g B = 1 1 B 3 Page 16

17 Hinged Beas... And we solve these to get: ( + ) g 1 = tan A A = 1 + g 1 A A ( + ) g 1 = tan B g B = 1 B 1 B 33 ecture 6, Act 3 Statics In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the sae ass and length. (a) case 1 (b) case (c) sae T 1 T / case 1 case 34 Page 17

18 ecture 6, Act 3 Solution Consider the total torque about an ais through the hinge: In both cases the CM of the strut is in its center, and the torque due to gravit is the sae = g /. The torque due to the wire ust also be the sae in both cases. T 1 T / g g case 1 case 35 ecture 6, Act 3 Solution In case 1 the torque due to the wire is τ 1 = T 1 sin(30) In case the torque due to the wire is τ = T sin(30) / In order for these to be the sae T = T 1 T 1 T / g g case 1 case 36 Page 18

19 ecture 6, Act 4 Statics A bo is placed on a rap in the configurations shown below. Friction prevents it fro sliding. The center of ass of the bo is indicated b a blue dot in each case. In which cases does the bo tip over? (a) all (b) & 3 (c) 3 onl ecture 6, Act 4 Solution We have seen that the torque due to gravit acts as though all the ass of an object is concentrated at the center of ass. Consider the botto right corner of the bo to be a pivot point. If the bo can rotate in such a wa that the center of ass is lowered, it will! Page 19

20 ecture 6, Act 4 Solution We have seen that the torque due to gravit acts as though all the ass of an object is concentrated at the center of ass. Consider the botto right corner of the bo to be a pivot point. If the bo can rotate in such a wa that the center of ass is lowered, it will! Page 0