1. Fu = ma NOTES p.19 DYNAMICS Recall that mass (kg) acceleration (m/s2)
Unbalanced
Force (N)
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!

2. Fu = F1 + F2 + F3 + ….
At SG level we used to subtract to find the unbalanced force …
100N
40N
We would say Fu = 100 – 40 = 60N
At H level we should use a sign convention …
100N
40N - +
So we say Fu = F1 + F2 = (-40) = +60N

3. - + Ff = ? Fe = 3000 N Example 1 – Friction on a horizontal plane.
A car’s engine force is 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
Fe = 3000 N
Fu = ? +ve
a = ? +ve
2nd Find “a”
a = v-u t
u = 0 ms-1
v = 18 ms-1
a = ?
t = 14 s = 14
= 1.3 ms-2

4. 3rd Calculate Fu
Fu= m a
Fu= ?
m = 900 kg
a = 1.3 ms-2
= 900 x 1.3
= 1170 N
4th Calculate Ff
Fu = Fe + Ff
Ff = Fu Fe
Ff =
Ff = N
So force of friction is 1830 N in opposite direction to car.

5. 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)
m = W g a)
W= 9800 N
m = ?
g = 9.8 Nkg-1
= 9800
9.8
= 1000 kg

6. b) What’s the cable tension as the lift moves down at 2 ms-1?
Tension is still 9800 N as constant velocity requires forces to
be balanced.
Determine the cable tension as the lift decelerates at
1.5 ms-2 while moving down? c)
1st a = ms-2
+ve
m = 1000 kg
T =?
W = N
2nd Fu = m a
= 1000 x 1.5
Fu = ?
a = 1.5 ms-2
= N
3rd Fu = T + W
T = Fu - W
T = 1500 – (-9800)
= N

7. Problems 52 – 65
4900 N
a) (i) m/s2 (ii) 3 x 106 N b) 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.
0.02 m/s N a) 120 N b) -20 N
a) N b) 108 m c) N
a) (ii) 7.7 m/s 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) m/s 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.
a) kg b) N N
a) N b) N c) N d) N
a) (i) N (ii) N (iii) N (iv) N b) 4.2 m/s2

8. Problems 52 – 65 (cont.)
63. c) An empty lift so resultant force would increase giving a higher acceleration.
51.2 N
b) N, N, N.

9. Example 3 – Towed Objects.
A car tows a caravan as shown
1200 kg
1000 kg
Friction
against car = 200 N
Friction
against caravan = 500 N
The car accelerates at 2 ms-2 (left!)
What’s the engine force of the car?
What force does the towbar exert on the caravan?

10. a) Free body diagram
-ve
+ve
Total mass
= 2200 kg
Fe = ?
Ff = -700 N
a = + 2 ms-2
Fu = ?
1st Fu = ma
= 2200 x (+2)
= N
2nd Fu = Fe + Ff
Fe = Fu - F f
Fe = – (-700)
= 5100 N

11. b) What force does the towbar exert on the caravan?
is -ve
is +ve
Caravan only
m = 1000 kg
Ftowbar = ?
Ff = N
a = + 2 ms-2
Fu = ?
1st Fu = ma
= 1000 x 2
= N
2nd Fu = Ftowbar F f
F towbar = Fu F f
= (-500)
= N

12. Problems 51 – 70.