NOTES p.19
Recall that
DYNAMICS
Fu = ma
Unbalanced
Force (N)
mass
(kg)
acceleration
(m/s2)
Therefore, the Newton is the unbalanced force which will
cause a mass of 1 kg to have an acceleration of 1 m/s2
Free body diagrams show the direction of all forces acting
on one point.
REMEMBER to use the + and – sign convention!
Fu = F1 + F2 + F3 + ….
At SG level we used to subtract to find the unbalanced
force …
40N
100N
We would say Fu = 100 – 40 = 60N
At H level we should use a sign convention …
-
40N
100N
+
So we say Fu = F1 + F2 = 100 + (-40) = +60N
Example 1 – Friction on a horizontal plane.
A car’s engine force is 3 000 N. If the mass is 900 kg and
it accelerates from rest to 18 ms-1 in 14 s, what is the force
of friction?
1st
Diagram + sign convention
Ff = ?
+
900 kg
Fu = ?
a=?
2nd Find “a”
u = 0 ms-1
v = 18 ms-1
a=?
t = 14 s
Fe = 3000 N
+ve
+ve
a = v-u
t
= 18 - 0
14
= 1.3 ms-2
3rd Calculate Fu
Fu= ?
m = 900 kg
a = 1.3 ms-2
Fu= m a
= 900 x 1.3
= 1170 N
4th Calculate Ff
Fu = Fe + Ff
Ff =
Fu
-
Fe
Ff =
1170 - 3000
Ff = - 1830 N
So force of friction is 1830 N in opposite direction
to car.
Example 2 – Lift Cable Tension.
A lift cable has a tension of 9800 N when the lift is at rest.
a) Determine the mass of the lift?
T = 9800 N
m=?
W = 9800 N (as W balances T)
a)
W= 9800 N
m=?
g = 9.8 Nkg-1
m=W
g
= 9800
9.8
= 1000 kg
b) What’s the cable tension as the lift moves down at 2 ms-1?
b)
Tension is still 9800 N as constant velocity requires forces to
be balanced.
c) Determine the cable tension as the lift decelerates at
1.5 ms-2 while moving down?
c)
+ve
1st a = + 1.5 ms-2
T =?
Fu = ?
m = 1000 kg
W = - 9800 N
a = 1.5 ms-2
2nd Fu = m a
= 1000 x 1.5
= + 1500 N
3rd Fu = T + W
T = Fu - W
T = 1500 – (-9800)
= + 11300 N
Problems 52 – 65
52. 4900 N
53. a) (i) 0.015 m/s2 (ii) 3 x 106 N b) -0.0027 m/s2
54. OA … decreasing acceleration as air resistance increases
AB…constant velocity as air resistance balances the weight
BC … parachute opens so dramatic deceleration to lower speed
CD … constant velocity as new air resistance balances weight
DE …dramatic deceleration as parachutist lands.
55. 0.02 m/s
56. 150 N
57. a) 120 N b) -20 N
58. a) 1200 N
b) 108 m
c) 2592 N
59. a) (ii) 7.7 m/s2
b) mass decreases as fuel is burned, air resistance decreases
as rocket leaves Earth’s atmosphere, weight decreases as rocket leaves Earth’s
gravitational field.
c) 15.9 m/s2
d) Once out of Earth’s gravitational pull weight drops to
zero so engines can be switched off and Earth will continue at a constant
velocity.
60. a) 1778 kg b) 62 424 N
62. a) 1960 N
b) 2260 N
63. a) (i) 2450 N
(ii) 2450 N
61. 28 600 N
c) 1960 N
(iii) 2950 N
d) 1660 N
(iv) 1950 N
b) 4.2 m/s2
Problems 52 – 65 (cont.)
63. c) An empty lift so resultant force would increase giving a higher acceleration.
64. 51.2 N
65. b) 37.2 N, 39.2 N, 43.2 N.
Example 3 – Towed Objects.
A car tows a caravan as shown
1200 kg
Friction
against car = 200 N
1000 kg
Friction
against caravan = 500 N
The car accelerates at 2 ms-2 (left!)
a) What’s the engine force of the car?
b) What force does the towbar exert on the caravan?
a) Free body diagram
+ve
Fe = ?
-ve
Total mass
= 2200 kg
a = + 2 ms-2
Fu = ?
1st
Fu = ma
= 2200 x (+2)
= +4400 N
2nd
Fu = Fe + Ff
Fe = Fu - F f
Fe = +4400 – (-700)
= 5100 N
Ff = -700 N
b) What force does the towbar exert on the caravan?
Ftowbar = ?
is -ve
is +ve
Caravan only
Ff = - 500 N
m = 1000 kg
a = + 2 ms-2
Fu = ?
1st
Fu = ma
= 1000 x 2
= + 2000 N
2nd
Fu
= Ftowbar + F f
F towbar = Fu
- Ff
= 2000 - (-500)
= + 2500 N
Problems 51 – 70.