Structured-type questions (4) HKASL Physics Examination format and allocation of marks Paper Format Weighting 1 (1 hour 10 min.) Structured-type questions (4) 33% 2 (1 hour 50 min.) MC (25) 30% Essays (2 out of 4) 22% 3 (2 years) TAS 15% Manhattan Press (H.K.) Ltd. © 2001

1.1 Statics 靜力學 and Dynamics 動力學 Manhattan Press (H.K.) Ltd. © 2001

What is statics? Statics is the branch of physics concerned with the analysis of loads (force, torque/moment) on physical systems in equilibrium. When in equilibrium, the forces in the system will have zero resultant and zero turning effect which will not cause any change in the motion of the object. Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Zero resultant F1 If the block is in equilibrium, the resultant force acting on it is zero. Resolve horizontally, F1 cos q = F2 Resolve vertically, F1 sin q = F3 q F2 F3 Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Vector approach F1 q F2 F3 If an equilibrium system consists of n forces, Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Vector Addition Triangular law Parallelogram law Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Polygon law Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Zero resultant forces F4 F3 F5 F6 F2 F1 Manhattan Press (H.K.) Ltd. © 2001

Turning effect of a force ─ rotate about axes 12.1 Turning effect of a force (SB p.170) Turning effect of a force ─ rotate about axes Pivot (or fulcrum) ─ position of axes axis pivot Manhattan Press (H.K.) Ltd. © 2001

Moment ─ the turning effect of a force 12.1 Turning effect of a force (SB p.170) Moment Moment ─ the turning effect of a force Moment arm ─ perpendicular distance between the force and the pivot moment arm Moment ＝Force  Moment arm ＝F  d door hinge (pivot) Unit of moment: N m door hinge Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 12.1 Turning effect of a force (SB p.171) Moment Moment Moment of F1 = F1  d1 Moment of F2 = F2  d2 Same turning effect F1  d1 = F2  d2 As d1 > d2, 　　F1 < F2 door hinge Note: the longer d, the smaller will be the force required. Manhattan Press (H.K.) Ltd. © 2001

Which requires a smaller force ? (d2 > d1) 12.1 Turning effect of a force (SB p.172) Moment Which requires a smaller force ? (d2 > d1) force force pivot  case 1 case 2 Manhattan Press (H.K.) Ltd. © 2001

Moment ─ clockwise or anticlockwise 12.1 Turning effect of a force (SB p.172) Moment Moment ─ clockwise or anticlockwise an anticlockwise moment a clockwise moment pivot anticlockwise clockwise Manhattan Press (H.K.) Ltd. © 2001

Principle of moments Moment of F1 = 10  0.4 = 4 N m (anticlockwise) 12.2 Principle of moments (SB p.174) Principle of moments Moment of F1 = 10  0.4 = 4 N m (anticlockwise) Moment of F2 = 5  0.8 = 4 N m (clockwise) pivot Two moments： same in magnitude, but in opposite direction cannot turn Manhattan Press (H.K.) Ltd. © 2001

Take the mid-point of the ruler as the pivot 12.2 Principle of moments (SB p.175) Take the mid-point of the ruler as the pivot = 1.6  10  0.1 + 1  10  0.4 = 5.6 N m Total clockwise moment pivot Manhattan Press (H.K.) Ltd. © 2001

Take the mid-point of the ruler as the pivot 12.2 Principle of moments (SB p.175) Take the mid-point of the ruler as the pivot = 0.4  10  0.2 + 1.2  10  0.4 = 5.6 N m Total anticlockwise moment pivot Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 12.2 Principle of moments (SB p.175) Principle of moments anticlockwise moment clockwise moment When a body is in balance, Total clockwise moment ＝Total anticlockwise moment Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 12.2 Principle of moments (SB p.176) Class Practice 1：Edmond, Jessie and Tracy are sitting on a seesaw at the positions shown in the figures. Given that their masses are 65 kg, 40 kg and 50 kg respectively, and the mass of the seesaw is negligible. Find the distance of Jessie from the pivot (d) when the seesaw is balanced. 400 N 500 N 650 N pivot d 1 m 1.7 m Apply the principle of moment, Clockwise moment = Anticlockwise moment 650 x 1.7 = 500 x 1 + 400 x d d = 1.51 m Answer Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Find the reaction from the pivot. R 400 N 500 N 650 N pivot d 1 m 1.7 m Since the system is in equilibrium, the resultant is zero. R = 400 + 500 + 650 = 1550 N Manhattan Press (H.K.) Ltd. © 2001

Take moment about other points R = 60 N 0.5m 0.2m 0.2m X Y 10 N 20 N Pivot O 30 N Take moment about X Clockwise moment Anti-clockwise moment = 30 x 0.7 + 10 x 0.9 = 30 Nm = 60 x 0.5 = 30 Nm Manhattan Press (H.K.) Ltd. © 2001

Take moment about other points R = 60 N 0.5m 0.2m 0.2m 0.1m X Y 10 N 20 N Pivot O 30 N Take moment about Y Clockwise moment Anti-clockwise moment = 60 x 0.5 = 30 Nm = 10 x 0.1 + 30 x 0.3 + 20 x 1 = 1 + 9 + 20 = 30 Nm Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Conclusion When an object is in equilibrium, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about that point. Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Two blocks are placed on a light metre ruler as shown. Find the reactions at X and Y. 3 kg 2 kg X Y 0.25 m 0.75 m Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Step 1: Draw all forces acting on the metre ruler (in equilibrium). 3 kg RX 2 kg RY X Y 20 N 30 N 0.25 m 0.75 m Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Step 2: Apply the principle of moment for objects in equilibrium. 3 kg RX 2 kg RY X Y 20 N 30 N 0.25 m 0.75 m Take moment about X (any point), 20 x 0.25 + 30 x 0.75 = RY x 1 RY = 27.5 N Resolve vertically, RX + RY = 20 + 30 RX = 50 – 27.5 = 22.5 N Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 More on moment of force Find the moment of F about O. q F d sin q Pivot O q d F Moment of F about O = Force x perpendicular distance from O = F x d sin q = Fd sin q Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 More on moment of force Find the moment of F about O. (Alternative method) F cos q F sin q Pivot O q d F Moment of F about O = Force x perpendicular distance from O = F sin q x d = Fd sin q Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 Example 2 A sign of mass 5 kg is hung from the end B of a uniform bar AB of mass 2 kg. The bar is hinged to a wall at A and held horizontal by a wire joining B to a point C which is on the wall vertically above A. If angle ABC = 30o, find the tension in the wire and the reaction exerted from the hinge. C Take moment about A, (Tsin 30o)(2d) = (20)(d) + (50)(2d) T = 120 N Resolve vertically, Ry + Tsin 30o = 20 + 50 Ry = 10 N Resolve horizontally, Rx = T cos 30o Rx = 103.9 N R = sqrt(Rx2 + Ry2) = 104.4 N q = tan -1(Ry/Rx) = 5.50o T T sin 30o T cos 30o Ry q R 30o A Rx d hinge 20 N 50 N Tension is 120 N and the reaction from the hinge is 104.4 N at an angle of 5.50o to the horizontal. Manhattan Press (H.K.) Ltd. © 2001

Moment and couple Couple ─ consists of 2 equal and opposite parallel forces whose lines of action do not coincide (重疊). F F d/2 d/2 d F torque of couple = F x d/2 + F x d/2 = Fd Manhattan Press (H.K.) Ltd. © 2001

Short test after each chapter Homework (SQ 2, 5, LQ 1) Next day 1 Short test : Moment Manhattan Press (H.K.) Ltd. © 2001

Manhattan Press (H.K.) Ltd. © 2001 The End Manhattan Press (H.K.) Ltd. © 2001