2 Force and MotionChapter 5 Moment of a Force
5Moment of a Force
Practice 5.1 (p.185)
1A
Moment about X
= Fd = 30 (0.7) sin 15 = 5.44 N
2D
Maximum magnitude of moment
= FD = 5 (0.2 2) = 2 N m
3A
4D
Just before the nail moves,
clockwise moment
= anticlockwise moment
100 0.3= f 0.05
f= 600 N
5D
6The moment arms are the same for the three forces.
They produce moments of the same magnitude.
7(a)Moment about O
= 5 0.25 cos 35
= 1.02 N m (anticlockwise)
(b) /= cos 30
OX == 1.386 m
Moment about O
= 50 OX sin 65
= 50 1.386 sin 65
= 62.8 N m (anticlockwise)
8(a) /(b)Normal reaction N acting on P by the plank and the forceFPacting on the plank by P
(c)Since P is stationary,
N = W
By Newton’s third law,
N = FP
FP= W
(d)Clockwise moment
= anticlockwise moment
M 2d= md
M=
The mass of Q is .
Practice 5.2 (p.198)
1B
Take moment about the pivot.
Sum of clockwise moment
= sum of anticlockwise moment
600 3= 400 (1 + 2) + W 2
W= 300 N
2D
Distance of c.g. from pivot = 10 = 15 cm
Distance of P from pivot
= 50 10 15 = 25 cm
Take moment about the pivot.
Sum of clockwise moment
= sum of anticlockwise moment
20 15=F 25 sin 40
F= 18.7 N
3C
4A
5After releasing, the c.g. of the ruler is outside the base of support (the table). Therefore, the ruler loses balance and falls.
6No, he cannot. When he stands against a wall, he cannot move backwards and his centre of gravity would be outside the base of support, i.e. his feet, if he bends over.
7 /When the roly-poly is tilted as shown, its weight produces a clockwise moment about the contact point and makes it return to the original position. Therefore, it stays upright.
8(a)The c.g. of the bus will be higher when passengers stand on the upper deck. The c.g. may be shifted outside the base of support even if the bus is only tilted slightly. This produces a net moment and makes the bus topple over.
(b)If the c.g. is not vertically below the point of suspension, the weight of the object will produce a moment that makes the object rotate. This moment will be zero only if the c.g. is vertically below the point of suspension.
Revision exercise 5
Cocept traps (p.200)
1F
A couple, which consists of two equal and opposite forces, can make an object rotate.
2T
Multiple-choice questions (p.200)
3A
4B
Magnitude of moment
= FD
= 8 0.4 sin 130
= 2.45 N m
5C
The c.g. of the rod is at its middle.
Take moment about X.
Clockwise moment= anticlockwise moment
W0.5= 20 1
W= 40 N
6B
7D
8B
The force acting on the rod by the pivot at O is not zero.
(1) is incorrect.
Take moment about O. Also, take clockwise moment as positive.
Net moment= 6 0.07 10 0.02 4 0.02
= 0.14 N m
(2) is correct.
The rod is not in equilibrium. The net moment about different points may be different.
(3) is incorrect.
9(HKALE 2008 Paper 2 Q1)
10(HKALE 2009 Paper 2 Q2)
11(HKDSE 2012 Paper 1A Q6)
12(HKDSE 2014 Paper 1A Q3)
Conventional quesionts (p.201)
13(a) /(1 correct force with correct name)1A
(All correct)1A
(b)No, it cannot.1A
If the c.g. is not vertically above Y, itis outside the base of support and X will lose balance.1A
14(a)Just before toppling, the force by the table will act on the stand at A.
Take moment about A.
Clockwise moment
= anticlockwise moment
2.4 9.81 3=m 9.81 101M
m= 0.72 kg1A
The maximum mass is 0.72 kg.
(b)Shorten the distance between the clamp and the stand.1A
Put a weight on the stand.1A
(Or other reasonable answers)
(c)When the system is put on an inclined plane, the horizontal distance d1between X and A will be shorterand the horizontal distance d2between the weight and A will be longer. 1A
Since m =,1A
the answer to (a) will be smaller.1A
15(a)(i)Take moment about the contact point.
Anticlockwise moment
= clockwise moment
F(5) = 1.5(9.81)15 + 5(9.81)30
1M
= 338 N1A
The force exerted by the biceps is 338 N.
(ii)No,1A
there is an inclined force acting on the forearm at the contact point.1A
(b)The weight is further away from the pivot (the shoulder).1A
Therefore, the clockwise moment produced by the weight about the pivot is larger.1A
He must exert a larger force on the arm to produce a larger anticlockwise moment.1A
As a result, he feels more tired.
16(a) /(1 correct force with correct name)1A
(All correct)1A
(b)Consider the vertical direction.
Since the board is stationary, by Newton’s first law,
2T sin 50= 10 1M
T= 6.53 N1A
The tension in each string is 6.53 N.
(c)(i) /Let the length of PQ be 2d, the length of SX be c and the weight of the mass be W.
Take moment about P.
Wc + 10d= T2(2d) sin 501A
T2=
Take moment about Q.
W(2dc)+ 10d= T1(2d) sin 50
T1=
T21A
The left string breaks.1A
(ii) /(C vertically below M)1A
17(a)No,1A
this is because the two forces point in the same direction.1A
(b)The force acting on the woman’ shoulder
= 200 + 200 = 400 N1A
(c)It would be the same.1A
(d)The heavier load should be put closer to her.1A
(e)She has to apply a downward force on the pole to keep it in equilibrium.1A
The total force acting on her shoulder would become larger.1A
Therefore, it would be harder if the two loads are put together.1A
18(a)Since the plank does not move, by Newton’s first law,
TB + 110= 50 + 150
TB= 90 N1A
The tension is 90 N.
(b)Take moment about the contact point between A and the plank.
Anticlockwise moment
= clockwise moment
90 65= 50d + 150
1M
d= 12 cm1A
The distance between the robot and string A is 12 cm.
(c)(i)Take moment about the contact point between B and the plank.
Anticlockwise moment
= clockwise moment
50(65 d) + 150(50 20)
= 115 651M
d= 5.5 cm1A
The minimum distance is 5.5 cm.
(ii)No, it cannot.1A
Take moment about the contact point between string B and the plank. The anticlockwise moment will be larger when the robot is on the left of A than when it is on the right. 1A
Therefore, the tension needed by Awill be even larger1A
and A will break.
19(a)Take moment about the pivot.
Clockwise moment
= anticlockwise moment
500(9.81)12= (m + 300)(9.81)61M
m= 700 g1A
The mass of the object is 700 g.
(b)Since the pan and the 500-g mass have the same acceleration,1A
the result will remain unchanged.1A
(c)Use a heavier mass to replace the 500-g mass.1A
Move the pivot closer to the pan.1A
20(a)Let T be the tension in the cable and N be the normal reaction.
Take moment about Q.
Clockwise moment
= anticlockwise moment
(Tsin 60)(1.6 + 0.7)
= 9000 0.7 + (Tcos 60)0.4
1M
T= 3516 N
3520 N1A
Consider the vertical direction.
3516 sin 60 + N= 90001M
N= 5960 N1A
The tension in the cable is 3520 N and the normal reaction is 5960 N.
(b)(i)No1A
(ii)Yes1A
21(a)The centre of gravity of a rigid body is the position that the weight of the body seems to act.1A
(b)No net force acts on it.1A
No net moment acts on it.1A
(c)(i)Take moment about X.
Clockwise moment
= anticlockwise moment
400(0.9) sin 40= T(0.6) sin 70
1M
T= 410.4 N
410 N1A
The tension in the rope is 410 N.
(ii) /Consider the vertical direction.
400 = 410.4 sin 20 + F sin 1M
F sin = 259.6(1)
Consider the horizontal direction.
Fcos = 410.4 cos 20(2)1M
(1) (2),
tan = 0.673
= 33.94
33.91A
From (2),
F == 465 N1A
The required force is 465 N (towards right at 33.9 above horizontal).
22(HKALE 2009 Paper 1 Q2(a))
23(HKALE 2012 Paper 1 Q2)
Experiment questions (p.205)
24Set up the apparatus as shown.
(Correct set-up)1A
Pull the spring balance as shown so that the ruler remains horizontal.1A
Measure the force Fwith the spring balance and the angle with the protractor.1A
Record several sets ofF and while keeping the ruler horizontal.1A
Plot a graph of F against . If the graph is a straight line passing through the origin, F is inversely proportional to sin . 1A
Physics in article (p.205)
25(a)(i)Moment about T
= 3(0.09 2) sin 451M
= 0.382 N m (clockwise)1A
(ii)Let W be the bowl’s weight, r be the bowl’s radius, F be the force acting along the slanted surface and be the slanting angle.
Take moment about T.
Moment produced by F
= F(2r) sin = 2Fr sin
Moment produced by W= Wr
To avoid toppling,
Wr 2Fr sin
11A
is independent of the radius r and increases when decreases.1A
Therefore, decreasing helps stabilize the bowl while increasingr does not.
(b)Zero1A
New Senior Secondary Physics at Work (Second Edition)1
Oxford University Press 2015