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
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
RN
Since the box remains at rest
30 + 35 – R = 0
So
R = 65 N
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 = 25
v = 35
Required equation: v = u + at
D – 400 = 1000 0.5
t = 20
a?
35 = 25 + 20a  a = 0.5 ms-1
 D = 900 N
Driving force = 900 N
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 490 . Find
the acceleration of the skip.
a ms-2
300 N
350 N
400 kg
490 N
Net force = 300 + 350 – 490 = 160 N
F = ma
160 = 400 a
a = 0.4 ms-2
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 1800. By modelling the car as a particle, find the
magnitude of the resistance it is experiencing.
2 ms-2
800 kg
1800 N
Net force = 1800 – R
1800 – R = 800 a
R=
200 N
RN
F=ma
= 800  2
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
0 ms-2
Net force = P + P + 10 - 100
PN
Constant velocity
P + P + 10 – 100 =
2P – 90 = 0
P=
45 N
(P+10) N
50 kg
0
100 N
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
200 kg
150 N
T – 150 = 200 a
TN
Data: u = 0
t =5 s = 30
a?
Required equation: s = ut + ½ at2
30 = ½  a  52
T – 150 = 200  2.4
Gives T = 630 N
gives: a = 2.4