Announcements Civil and echanical engineers: his week is for you! Ø Sta;cs: Oooh, so exci;ng! Ø Please pay aaen;on: We want you to build bridges that don t fall down! Exam 3 next Wednesday!!! (November 18) Ø Covers units 11 17 (conserva;on of momentum sta;cs) Ø Same rules as always Ø Will post exam scenarios Wednesday (prac;ce exam already posted) echanics ecture 17, Slide 1
ecture 17: Rotational Statics part 1 oday s Concepts: a) orque Due to Gravity b) Sta;c Equilibrium echanics ecture 17, Slide
New opic, Old Physics: Sta;cs: As the name implies, sta;cs is the study of systems that don t move. adders, sign- posts, balanced beams, buildings, bridges... he key equa;ons are familiar to us:! F =! ma!! τ = Iα If an object doesn t move: ~a =0 ~ =0 F! = 0 τ! = 0 he net force on the object is zero he net torque on the object is zero (about any axis) echanics ecture 17, Slide 3
Statics: Example: What are all of the forces ac;ng on a car parked on a hill? y x N f θ mg echanics ecture 17, Slide 4
Car on Hill: Use Newton s nd aw: F NE = A C = 0 Resolve this into x and y components: x: f - mg sinθ = 0 f = mg sinθ y x y: N - mg cosθ = 0 N f N = mg cosθ θ mg echanics ecture 17, Slide 5
orque Due to Gravity R R! C agnitude: τ = R C g sin (θ ) = gr ever arm echanics ecture 17, Slide 6
Example: Consider a plank of mass suspended by two strings as shown. We want to find the tension in each string: 1 x cm / /4 g y x echanics ecture 17, Slide 7
Balance forces: F! = 0 1 + = g 1 x cm / /4 g y x echanics ecture 17, Slide 8
Balance orques τ! = 0 Choose the rota;on axis to be out of the page through the C: he torque due to the string on the right about this axis is: τ = /4 he torque due to the string on the le^ about this axis is: τ 1 = - 1 / 1 + x cm / /4 g Gravity acts at C, so it exerts no torque about the C y x echanics ecture 17, Slide 9
Finish the problem he sum of all torques must be 0: τ +τ 0 + 0 1 = = 1 4 1 = 1 We already found that 1 + = g x cm / /4 1 = = 1 g 3 g 3 g y x echanics ecture 17, Slide 10
What if you choose a different axis? τ! = 0 Choose the rota;on axis to be out of the page at the le^ end of the beam: he torque due to the string on the right about this axis is: τ = 3 4 he torque due to the string on the le^ is zero τ 1 = 0 orque due to gravity: + τ g = g 1 x cm / /4 g y x echanics ecture 17, Slide 11
You end up with the same answer! he sum of all torques must be 0: τ + τ + τ = 0 1 g g = 0 = 3 g 3 4 1 We already found that 1 + = g x cm 1 = = 1 g 3 g 3 / /4 g Same as before y x echanics ecture 17, Slide 1
Approach to Statics: Summary In general, we can use the two equa;ons F! = 0! τ = 0 to solve any sta;cs problem. When choosing axes about which to calculate torque, we can some;mes be clever and make the problem easier... echanics ecture 17, Slide 13
CheckPoint: Ball on Rod In Case 1 one end of a horizontal massless rod of length is aaached to a ver;cal wall by a hinge, and the other end holds a ball of mass. In Case the massless rod holds the same ball but is twice as long and makes an angle of 30 o with the wall as shown. In which case is the total torque about the hinge biggest? A) Case 1 B) Case C) Both are the same About 50% of the class got this right. A C B D gravity 90 o 30o Case 1 Case echanics ecture 17, Slide 14
AC: Rods, part I In Case 1 one end of a horizontal rod of mass and length is aaached to a ver;cal wall by a hinge, and the other end is free to rotate. In Case a rod of mass and length makes an angle of 30 o with the wall as shown. A C B D In which case is the torque about the hinge largest? A) Case 1 B) Case C) Both are the same gravity 90 o 30o Case 1 Case echanics ecture 17, Slide 15
AC: Rods, part II In Case 1 one end of a horizontal rod of mass and length is aaached to a ver;cal wall by a hinge, and the other end is free to rotate. In Case a rod of mass and length makes an angle of 30 o with the wall as shown. A C B D Which rod has larger angular accelera;on at the moment shown? A) Case 1 B) Case C) Both are the same gravity 90 o 30o Case 1 Case echanics ecture 17, Slide 16
AC: Ball and Beam An object is made by hanging a ball of mass from one end of a plank having the same mass and length. he object is then pivoted at a point a distance /4 from the end of the plank suppor;ng the ball, as shown below. How far to the right of the pivot is the center of mass of the plank only? A C B D A) /4 B) / C) 3/4 gravity /4 echanics ecture 17, Slide 17
CheckPoint: Ball and Beam An object is made by hanging a ball of mass from one end of a plank having the same mass and length. he object is then pivoted at a point a distance /4 from the end of the plank suppor;ng the ball, as shown below. Is the object balanced? A) Yes B) No, it will ;p le^ C) No, it will ;p right A C B D gravity /4 echanics ecture 17, Slide 18
CheckPoint: Sign In Case 1, one end of a horizontal plank of mass and length is aaached to a wall by a hinge and the other end is held up by a wire aaached to the wall. In Case the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire biggest? A) Case 1 B) Case C) Both are the same About 40% of the class got this right. gravity / Case 1 Case echanics ecture 17, Slide 19
AC: Sign Which equa;on correctly expresses the fact that the total torque about the hinge is zero? A C B D A) sin θ g = 0 B) g = 0 θ C) g sin θ = 0 g he cancels out: sinθ = g =) = g sin echanics ecture 17, Slide 0
CheckPoint: Revote In Case 1, one end of a horizontal plank of mass and length is aaached to a wall by a hinge and the other end is held up by a wire aaached to the wall. In Case the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire biggest? A) Case 1 B) Case C) Same A C B D / echanics ecture 17, Slide 1
ever Arm ever Arm: he perpendicular (shortest) distance between the line of ac,on of a force and the rota;on axis. r? = sin apple 90 r? = sin r? =/3sin echanics ecture 17, Slide
AC In which of the sta;c cases shown below is the tension in the suppor;ng wire bigger? In both cases is the same, and the blue strut is massless. A) Case 1 B) Case C) Same A C B D 1 30 0 30 0 / Case 1 Case echanics ecture 17, Slide 3
It s the same. Why? Case 1 Case 1 d 1 θ d θ / Balancing orques g = 1 d 1 = 1 sinθ g = d = 1 = g sinθ = g sinθ sinθ echanics ecture 17, Slide 4
mg cosθ g 1 cosθ H y H x Use hinge as axis τ sign + τ beam + τ wire = 0 1 mg cosθ + g cosθ sinθ = 0 3 sinθ 3 = 3m 3 + 4 g cotθ echanics ecture 17, Slide 5
mg y g x H y H x F x = 0 H x = 0 H x = F y = 0 H y mg g = 0 = ( m )g H y +! H = + H x H y echanics ecture 17, Slide 6
m = 3m + 3 4 g cotθ Solve for m echanics ecture 17, Slide 7
mg cosθ g H y Use hinge as axis τ sign + τ beam + τ wire = 0 1 mg cosθ + g cosθ sinθ = 0 3 sinθ 3 H x 1 cosθ Which of these things make smaller? echanics ecture 17, Slide 8