Rigid Body F=ma Fixed Axis Rotation: Example 3 A 64.4 lb rectangular plate is suspended by a pin and a cord. At the instant the cord is cut, please determine: (a) Pin reactions at A; (b) The plate’s angular acceleration, a . A 4 ft Cut! B G D A 3 ft C 64.4 lb Rectangular plate, released from rest. 2 ft 2.5 ft 1.5 ft G Draw a Free Body Diagram and a Kinetic Diagram Ay FBD KD 4 ft Ax G 3 ft = 2.5 ft 3 64.4 lb m = 64.4/32.2 = 2 slug 4 mat = m man = m 2 r =0 Since released from rest. r = 2 (2.5) =5 Calculate IG for the plate: [IG = (1/12)m(a2 + b2)] IG = (1/12)(2)(42+32)= 4.17 slug-ft2 Calculate Ipin from the parallel axis theorem: Ipin = IG + md2 = 4.17 + (2)(2.52) = 16.67 slug-ft2 Ay Ax 4 ft 0= m G 3 ft = 2.5 ft 3 64.4 lb 2r 4 mat = m r = 5 Write equations of motion: Sum moments about the pin: SMpin = Ipina ; (64.4)(2 ft) = Ipina = 16.67a a = 7.73 rad/sec2 from kinematics: at = ra = 2.5(7.73), at = 19.3 fps2 Note: Don’t be alarmed about the fact that we drew IGa on the KD and then used Ipin in the moment equation. Remember the Ipina term on the RHS is the equivalent of a kinetic moment of IGa plus mat(r) about the pin. Using Ipin simplifies the equation slightly. Ay Ax 4 ft 0= m G 3 ft = 2.5 ft 3 64.4 lb 2r 4 mat = 2at = 2 r = 5 Use force summations to find the pin reactions at A: SFy = may ; Ay – 64.4 = – 2(19.3)(4/5) Ay = 64.4 – 30.9 Ay = 33.5 lb SFx = max ; Ax = 2(19.3)(3/5) Ax = 23.2 lb As noted earlier, you may choose to express the pin reactions at A in normal and tangential components, too. Then sum forces in n and t to compute An and At. (The mg term would need to be resolved into n and t coordinates, though.) Ay Ax 4 ft 0= m G 3 ft = 2.5 ft 3 64.4 lb 2r 4 mat = 2at = 2 r = 5 If you choose to sum moments about the pin using a kinetic moment, (instead of using Ipina) you get: SMpin = IGa + mat(r); (64.4)(2 ft) = 4.17a + 2at(2.5) 128.8 = 4.17a + 5at ; but here sub in at = 2.5a 128.8 = 4.17a + 5(2.5a) = 16.67a a = 7.73 rad/sec2 etc. If you choose to sum moments about G, you introduce the pin reactions into the moment equation: SMG = IGa ; Ay(2) – Ax(1.5) = 4.17a (1) See the next page for the remainder of this approach…. Ay 4 ft 0= m Ax G 3 ft = 2.5 ft 3 64.4 lb 2r 4 mat = 2at = 2 r = 5 Write the F = ma equations: SFy = may ; Ay – 64.4 = – 2(at)(4/5) SFx = max ; Ax = 2(at)(3/5) You need a fourth equation: at = 2.5a. Solve these four equations for the four unknowns. 2 -1.5 0 -4.17 1 0 1.6 0 0 1 -1.2 0 0 0 1 -2.5 Ay 0 64.4 Ax = at 0 0 (2) (3) (4) Ay = 33.5 lb Ax = 23.2 lb at = 19.3 fps2 = 7.73 r/s2 These agree with our earlier results, but you can see that summing moments about G involves more solution effort. That’s why we recommend summing moments about the pin.