5 Forces
Practising problem solving solutions
Part 1
1 a) i) x: 20 cos 30 = 17.3 N
y: 20 sin 30 = 10 N
ii) x: 17 N →
y: 4 N ↓
2
2
2
b) i) F = 15 + 18 = 549; F = 23.4 N; tan θ = 15/18; θ = 39.8° to the horizontal
ii) Horizontally: 100 + 120 sin 45 = 185 N
Vertically: 150 + 120 cos 45 = 235 N
2
2
2
F = 185 + 235 = 89,450; F = 299 N; tan θ = 235/185; θ = 51.8° to the horizontal
iii) ΣF = A − 4 = 3 × 2; A = 10 N
iv) Force B = 30 N; ΣF = 4 × 10 = 40 N = Force A
−2
2 a) a = (5 – 0) / 20 = 0.25 m s ; ΣF = ma = 85 × 0.25 = 21 N
6
b) ΣF = T – mg = ma; T = 750 000 (9.81 + 2) = 8.9 × 10 N
2
2
2
−2
c) i) v = u + 2as; 0 = 7 + (2 × a × 20); a = −1.23 m s
ii) ΣF = 0.05 × 1.23 = 0.062 N
−2
3 a) 200 – (40 × g sin 25) = 40 a; a = 0.85 m s
b) R = 40 × g × cos 25 = 356 N
4 a) Moment = 10 × 3 = 30 N m, clockwise
b) Moment = 6 × 0.5 = 3 N m, clockwise
c) Moment = 0
d) Sum of moments = (4 × 3) + (10 × 3.5) = 47 N m, anti-clockwise
e) Sum of moments = (5 × 2) + (7 × 2) = 24 N m, anti-clockwise
f) Sum of moments = 10 × 2.5 = 25 N m, clockwise
Part 2
1 a) Horizontally: 20 cos 45 = 20 cos 45
Vertically: 2 × (20 sin 45) = 28 N
b) Horizontally: 240 cos 25 = 251 cos 30
Vertically: 240 sin 25 + 251 sin 30 = 227
−2
2 a) ΣF = (25 × g sin 30) – 10 = 25a; a = 4.5 m s
−2
b) ΣF = (2 × 25 × g sin 30) – 5 = (2 × 25) a; a = 4.8 m s
−2
3 a) AB: 30 / 600 = 0.05 m s
−2
BC: a = 10 / 600 = 0.017 m s
CD: a = 0
© Hodder & Stoughton Limited 2015
5 Forces
Practising problem solving solutions
5
b) AB: ΣF = (0.05 × 2 × 10 ) = 10 000 = tractive force – drag; 10 000 = T – 3000; T = 13 kN
5
BC: ΣF = (0.017 × 2 × 10 ) = 3400; 3400 = T – 3000; T = 6.4 kN
CD: 3 kN (forces are balanced)
2
c) D = 2000 + 20v + 3.5v ; D = 8400 N
4 a) 100 = F × 0.5 cos 30; F = 231 N
b) Moment of F = 231 × 0.5 = 115 N m
c) Moment of F = 231 × 0.5 cos 20 = 109 N m
d) Assuming F remains the same then the moment is smaller. To keep the same value of the
moment, F would have to be larger.
Part 3
1 Horizontally: X cos 25 = Y cos 45 + 45 cos 60; 0.91X – 0.71Y = 22.5
Vertically: X sin 25 + Y sin 45 = 45 sin 60 + 20; 0.42X + 0.71Y= 59
1.33X = 81.5; X = 61 N
Y = 47 N
2 a) Taking moments about O:
Anti-clockwise: 5 × 20 = 100 N cm
Clockwise: (5 × 15) + (F × 10) = 75 + 10F
For rotational equilibrium: 100 = 75 + 10F; F = 2.5 N
b) i) Taking moments about A:
Anti-clockwise: 12.5 × 20 = 250 N cm
Clockwise: (5 × 35) + (2.5 × 30) = 250 N cm 
ii) Taking moments about B:
Anti-clockwise: (5 × 35) + (2.5 × 5) = 187.5 N cm
Clockwise: 12.5 × 15 = 187.5 N cm 
3 a) F = (m + 4m) a; a = F/5m
b) FA = m × F/(5m) = F/5
c) FA = F – (force on A due to B); force on A due to B = F – FA = 4F/5
d) FB = –FA from Newton III
−1
2
2
2
−2
4 a) 60 mph = 26.7 m s ; v = u + 2as; 0 = 26.7 + (2 × 55 × a); a = – 6.46 m s
b) ΣF = 1800 × 6.46 = 11.6 kN
−2
c) m = 2160 kg; a = 11 600 / 2160 = 5.37 m s
2
0 = 26.7 + (2 × s × 5.37); s = 66.4 m
© Hodder & Stoughton Limited 2015