Forces and motion 10E
a R(→), F = (2 + 8) 0.4
=4
Hence F is 4 N.
b For Q:
R(→), T = 2 0.4
= 0.8
The tension in the string is 0.8 N.
c Treating the string as inextensible (i.e. it does not stretch) allows us to assume that the acceleration
of both masses is the same. Treating the string as light (i.e. having no/negligible mass) allows us to
assume that the tension is the same throughout the length of the string and that its mass does not
need to be considered when treating the system as a whole.
F = ma
a For the whole system: F = 60, m = 20 + m = 10, a = 2
60 = (20 + m) × 2
20 + m = 60 ÷ 2
m = 30 − 20
The mass of Q is 10 kg.
b For P: F = T, m = 20, a = 2
T = 20 × 2
The thrust in the rod is 40 N.
3 F = ma
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3 a For the whole system: F = 30, m = 8 + 7 = 15, a = ?
30 = 15a
The acceleration of the system is 2 m s−2.
b For P: F = T, m = 7, a = 2
T=7×2
The tension in the string is 14 N.
a Considering the system as a whole: total mass, m = 1700 + 110 + 190 = 2000 kg
Taking down as positive:
F = ma = mg − T
2000 × 1.8 = (2000 × 9.8) – T
T = 19600 – 3600
The tension in the cable is 16 000 N.
b i Force exerted on box A by box B is a normal reaction force, R1 which acts upwards.
For box A, taking down as positive:
110 × 1.8 = 110g − R1
R1 = 110(g −1.8)
R1 = 110 × 8
Box B exerts an upwards force of 880 N on box A.
ii Let downward force exerted on lift by box B be S.
For lift alone, taking down as positive:
1700 × 1.8 = 1700g + S – T
S = T + 1700(1.8 − g)
S = 16 000 − 13600 = 2400
Alternatively (or as check), use Newton’s third law of motion:
|Force exerted box B by box A| = |Force exerted on box A by box B| = 880 N
|Force exerted on lift by box B| = |Force exerted on box B by lift| = |R2|
For box B, taking down as positive:
190 × 1.8 = 880 + 190g – R2
R2 = 880 + 190(g −1.8)
R2 = 880 + 1520 = 2400
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5 F = ma
a For the whole system:
F = 50 000 – 10 000 − 4000 = 36 000
a=5
total mass = 3m + m = 4m
36 000 = a × total mass = 4m × 5 = 20m
m = 1800
so 3m = 5400
The mass of the lorry is 1800 kg, and that of the trailer is 5400 kg.
b For the trailer:
F = T – 10 000, m = 5400, a = 5
T – 10 000 = 5400 × 5 = 27 000
T = 37 000
The tension in the tow-bar is 37 000 N.
c Treating the tow-bar as inextensible (i.e. it does not stretch) allows us to assume that
the acceleration of the truck and the trailer are the same. Treating the tow-bar as
light (i.e. having no/negligible mass) allows us to assume that the tension is the
same throughout its length and that its mass does not need to be considered when
treating the system as a whole.
6 F = ma, W = mg
Taking upwards as positive
a For the whole system:
180 – 15g = 15a
15a = 180 – (15 × 9.8)
180 − 147
a= = 2.2
15
The acceleration is 2.2 m s−2.
b For B:
ma = T – W
5 × 2.2 = T – (5 × 9.8)
11 = T – 49
The tension in the string is 60 N.
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7 F = ma, W = mg
Taking up as positive
a For the whole system:
118 – (6 + m)g = (6 + m) × 2
118 = (6 + m)(2 + g) = (6 + m)(2 + 9.8)
118
= 6+m
11.8
10 = 6 + m
The mass of B is 4 kg.
b For B:
ma = T – W
4 × 2 = T – (4 × 9.8)
8 = T – 39.2
The tension in the string is 47.2 N.
8 F = ma
a For the whole system:
F = 12 000 – 2000 − R
m = 1600 + 6400 = 8000
a = 0.5
10 000 − R = 8000 × 0.5 = 4000
The resistance to the motion of the engine is 6000 N.
b For the carriage:
F = T – 2000, m = 1600, a = 0.5
T – 2000 = 1600 × 0.5 = 800
The tension in the coupling is 2800 N.
9 F = ma
a For the whole system:
F = 1200 – 1000 − 200 = 900
m = 900 + 300 = 1200
900 = 1200a
a = 900 ÷ 1200 = 0.75
The acceleration is 0.75 m s−2, as required.
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9 b For the trailer:
F = T – 100, m = 300, a = 0.75
T – 100 = 300 × 0.75 = 225
The tension in the towbar is 325 N.
Taking ← as positive
Deceleration = α
Force on trailer = resistance to motion + thrust from tow-bar
Using F = ma
100 + 100 = 300 α
200 2
= =
300 3
For car:
F + 200 −100 = 900α
2
F = 900 − 100 = 500
3
The force the brakes produce on the car is 500 N.
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