1. Force and motion Forces acting together
If several forces act on an object parallel to a given direction, then the net force is the sum of the forces in that direction minus the sum of forces in the opposite direction.
200 N
1200 N
Net force = 1200 – 200 = 1000 N
The net force is equal to the product of the mass of the object and its acceleration in given direction.
F = ma

2. Force and motion Example
H heavy box of mass 40 kg has a handle on one side. Two girls try to move it cross the floor. One pulls horizontally on the handle with a force of 30 N, the other pushes from the other side of the box with a force of 35 N, but the box does not move, find the frictional force resisting the motion.
35 N
30 N
40 kg
R N
Since the box remains at rest
– R = So
R =
65 N

3. Force and motion
A car of mass 1000 kg is moving with a constant speed of 25 m/s in a horizontal straight line, against a resisting force of 400 N. (a) What driving force is being provided to sustain this motion? (b) The driver speeds up uniformly over the next 20 seconds to reach a speed of 35 m/s. Assuming that the resisting force remains at 400 N, calculate the new driving force.
Driving force
400 N
(a)
No acceleration 
Driving force = 400 N
(b)
Data u = v = 35 t = 20 a?
Required equation:
v = u + at
35 = a
 a = 0.5 ms-1
D – 400 = 1000 0.5
 D = 900 N
Driving force = 900 N

4. Force and motion Example
Two builders push a skip of mass 400 kg across the ground. They both push horizontally, one with a force of 300 N, the other with 350 N. Motion is resisted by a frictional force of Find the acceleration of the skip.
a ms-2
300 N
400 kg
350 N
490 N
Net force =
– 490 =
160 N
F = ma
160 = 400 a
a =
0.4 ms-2

5. Force and motion Example
A car of mass 800 kg is travelling a straight horizontal road with an acceleration of 2 ms-2. The engine is exerting a forward force of magnitude By modelling the car as a particle, find the magnitude of the resistance it is experiencing.
2 ms-2
800 kg
1800 N
R N
Net force =
1800 – R
F = m a
1800 – R =
800 a
= 800  2
R =
200 N

6. Force and motion Example
A dinghy of mass 50 kg is moved across a horizontal beach at a steady speed of 2 ms-1. One of the crew pulls with a force P, the other pushes with a force of (P + 10) newtons. The frictional force resisting the motion is 100 N. Find P.
direction of motion
Net force =
P + P
0 ms-2
P N
(P+10) N
Constant velocity
50 kg
P + P + 10 – 100 =
100 N
2P – 90 = 0
P =
45 N

7. Force and motion Example
A wagon of mass 200 kg is pulled by a horizontal cable along a straight level track against a resistance force of 150 N. The wagon starts from rest. After 5 seconds it has cover a distance of 30m. Find the tension in the cable.
F = ma
a ms-2
T – 150 =
200 a
200 kg
T N
150 N
Data: u = 0 t =5 s = a?
Required equation: s = ut + ½ at2
30 = ½  a  52
gives: a =
2.4
T – 150 =
200 
2.4
Gives T =
630 N