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Quiz on moments. Term 3 Quiz 1. Question 1. 5.0 cm. 5.0 cm. F. 3.0 cm. 1.0 N. 3.0 cm. 3.0 N. (a) Point about which combined block has tendency to rotate. 5.0 cm. 5.0 cm. F. 3.0 cm. 1.0 N. 3.0 cm. 3.0 N. P. (b) Direction: Clockwise.
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Quiz on moments Term 3 Quiz 1
Question 1 5.0 cm 5.0 cm F 3.0 cm 1.0 N 3.0 cm 3.0 N
(a) Point about which combined block has tendency to rotate 5.0 cm 5.0 cm F 3.0 cm 1.0 N 3.0 cm 3.0 N P
(b) Direction: Clockwise Note: focus on the line of action of F and the position of the pivot. 5.0 cm 5.0 cm F 3.0 cm 1.0 N 3.0 cm 3.0 N P
(b) Direction: Clockwise Note: focus on the line of action of F and the position of the pivot. 5.0 cm 5.0 cm F 3.0 cm 1.0 N 3.0 cm 3.0 N P
(c) Anticlockwise moments Note: Each force produces one moment, which has a magnitude and a direction. 5.0 cm 5.0 cm CW F 3.0 cm ACW 1.0 N 3.0 cm 3.0 N ACW P
(c) Anticlockwise moments Consider the distance for each individual force. 5.0 cm 5.0 cm CW F 3.0 cm ACW 1.0 N 3.0 cm 3.0 N ACW P
(c) Anticlockwise moments For the first ACW force. 5.0 cm 5.0 cm 3.0 cm ACW Perpendicular distance for this force 1.0 N 3.0 cm ACW moment due to 1.0 N force = 1.0 N x 0.050 m = 0.050 Nm 5.0 cm P
(c) Anticlockwise moments For the second ACW force. 5.0 cm 5.0 cm 3.0 cm Perpendicular distance for this force 3.0 cm ACW moment due to 3.0 N force = 3.0 N x 0.050 m = 0.15 Nm 3.0 N 5.0 cm ACW P
(c) Anticlockwise moments Total ACW moments = 0.05 + 0.15 = 0.20 Nm 5.0 cm 5.0 cm 3.0 cm ACW 1.0 N 3.0 cm 3.0 N 5.0 cm ACW P
(d) Minimum F to start tilting the block Just before the block starts tilting, ACW moments about P = CW moments about P The only force which can produce a clockwise moment is F 5.0 cm 5.0 cm F 3.0 cm CW ACW 1.0 N 3.0 cm 3.0 N ACW P
(d) Minimum F to start tilting the block Just before the block starts tilting, ACW moments about P = CW moments about P 5.0 cm 5.0 cm F 3.0 cm CW Perpendicular distance for this force 3.0 cm 12.0 cm P
(d) Minimum F to start tilting the block Just before the block starts tilting, ACW moments about P = CW moments about P 0.20 Nm = F (0.12 m) 5.0 cm 5.0 cm F From (c) 3.0 cm CW ACW 1.0 N 3.0 cm Therefore F = 1.7 N 3.0 N ACW P
(a) Weight of uniform plank Weight = 3.0 N
(b) To find normal contact force at K Normal contact force Normal contact force Weight = 3.0 N Draw all forces acting on the plank first! Even if the question did not ask for it.
(b) To find normal contact force at K Normal contact force Normal contact force Weight = 3.0 N The plank is stationary The plank is in equilibrium Apply PoM
(b) To find normal contact force at K In order to apply PoM, (1) Choose a pivot Normal contact force Normal contact force Weight = 3.0 N Choose L since we don’t know the normal contact force at L.
(b) To find normal contact force at K (2) Identify CW and ACW moments produced by each individual force CW Normal contact force Normal contact force ACW Weight = 3.0 N ACW Sum of CW moments = Fk (3.0 m) Sum of ACW moments = (2.0 N)(2.0 m) + (3.0 N)(1.5 m)
(b) To find normal contact force at K (3) Equate the sum of CW and ACW moments CW Normal contact force Normal contact force ACW Weight = 3.0 N ACW Fk (3.0 m) (2.0 N)(2.0 m) + (3.0 N)(1.5 m)
(b) To find normal contact force at K CW Normal contact force Normal contact force ACW Weight = 3.0 N ACW Fk = 2.8 N (2 s.f.)
Bonus: Normal contact force at L Normal contact force Normal contact force Weight = 3.0 N The plank is in equilibrium No NET FORCE
Bonus: Normal contact force at L Normal contact force Normal contact force Weight = 3.0 N FL = 2.2 N (2 s.f.) Fk + FL = 2.0 N + 3.0 N
Quiz on moments (2) Term 3 Quiz 1 Requiz