Questions on moments
1. (i) pressure = force / area B1
(ii) moment = force multiplied by the perpendicular distance
(from the line of action of the force) to the pivot B1
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2. (a) The net force acting on the object must be zero B1
The net moment (about any point) must also be zero B1
(b) Taking moments about A, we have
Sum of clockwise moments = sum of anticlockwise moments C1
(0.25 × 200) + (5.0 × 9.81 × 0.4) = 0.8F C1
F = 87 (N) A1
(c) These forces are opposite but not equal in magnitude. B1
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3. (i) Moment is the force the perpendicular distance
from (the line of action of) the force to the pivot/point
(missing perpendicular 1, missing from the force to B2
the pivot / point 1)
(ii) Torque of a couple: one of the forces x B1
perpendicular distance between (the lines of action of) the forces
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4. (i) 1 3600 1.0 = X 2.5 C2
one mark for one correct moment, one mark for the
second correct moment and equated to first moment A0
2 X = 1440 (N) C1
Y = 3600 – 1440 or 3600 1.5 = Y 2.5 A1
= 2160 (N) B1
(ii) Not a couple as forces are not equal B1
and not in opposite directions / the forces are in the
same direction C1
(iii) P = F / A B1
= 1440 / 2.3 10–2 B1
= 62609 (6.3 104)
unit Pa or N m–2
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5. (i) 1 The (distribution of the) mass of the lawn mower is
not uniform B1
2. One correct moment about A stated
B 110 or 350 20 B1
B = (350 20) / 110 (moments equated) B1
B = 63.6 (N) A0
3. A = 350 63.6 = 286(.4) (N) A1
(ii) A goes down and B goes up B1
Turning effect of B is less / B needs greater force to
produce the same moment / if distance goes down
force needs to go up (to maintain the same turning effect) B1
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6. (a) (i) • F × 25 sin15 / F × 0.25 sin 15 for one moment. (1)
• 450 × 40 cos 30 / 450 × 0.4 cos 30 for the other moment. (1)
• moments equated or stated, even if not correct.
[Do not accept forces resolved vertically] (1)
• Answer F = 2409 (N). (1) 4
(ii) • Answer F = 951 (N). 1
(b) • Link large force (2409N) with small angle (30°) /
The more nearly horizontal / the smaller the angle with the horizontal
your back is, the greater the force needed (from the muscles). (1)
• the force is large because the anti-clockwise moment is large (1)
• the anti-clockwise moment is large because the perpendicular
distance to the pivot is large. (1)
(First 3 points + any one of the following:) (1)
• consequence, eg tendon ‘goes’, etc.
• (Therefore) keep your back as vertical / upright as possible,
• … with the load close to your body …
• … and bend your knees / use leg muscles to do some of the lifting.
• ..back is strong in compression / weak in shear, etc. 4
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7. (a) (i) force drawn vertically upwards at plunger B1
force drawn vertically at H B1
(ii) 20 500 / force on Plunger 120 (one correct moment stated) B1
Plunger force 120 = (20 500) B1
Plunger force = 83(.3) (N) A0
(b) (i) pressure = force / area
= 83 / 4 10–3 C1
= 20800 (Pa) A1
(ii) decrease area of plunger / decrease distance H to plunger /
increase F / increase length of arm B2
MAX 2
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8. (i) W vertically down at G B1
Force at O vertical B1
(ii) V × 0.9 × cos60 = W × 0.35 × cos60 B1
V = (25 × 0.35) / 0.9 B1
= 9.7(22) (N) A0
(iii) total force is zero stated or implied / 25 – 9.7 C1
force at hinge = 15.3 (N) A1
(or may take moments about G or V)
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