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Newton's Laws: More worked examples
Example 6
Iift, L
A glider weighing 8. 0 kN is being towed horizontally through the air
at a steady speed. The diagram shows the forces acting on the glider.
If the drag force D is 600 N, calculate :
a) the tension T in the tow-line,
b) the lift force L on the glider.
weight, W
a) Steady horizontal speed means no horizontal acceleration.
So resultant horizontal force = 0
total force acting to the right
.•. total force acting to the left
T cos35°
D = 600 N
600 N
= 732.5 N = 7.3 x 10 2 N (2 s.f.)
:. T
cos 35°
b) Travelling horizontally so there is no vertical acceleration
So resultant vertical force = 0
total force acting downwards
.•. total force acting upwards
L
W + T cos55°
8000 N + (730 N x cos 55°)
Components of T:
T cos35°
35°
___,T cos 55°
= 8420 N = 8.4 x 10 3 N
Example 7
(2 s.f.)
kd
The diagram shows the forces acting on a person in a stationary lift.
The person has a mass of 70 kg. Taking g = 10 N kg- 1 ,
calculate the contact force R when :
a) the lift is at rest,
b) the lift is accelerating upwards at 1. 0 m s- 2 ,
c) the lift is accelerating downwards at 2.0 m s- 2 ,
d) the lift is ascending at a steady speed.
- contact force R
of lift on person
•
a) As the person is not moving, the resultant force on him is zero .
weight W acting downwards
.•. contact force R acting upwards
mass X g
70 kg x 10 N kg- 1 = 700 N
b) Accelerating upwards. Applying Newton's second law:
mass X acceleration
resultant upwards force (R - W)
R- 700 N
(70 kg x 1.0 m s- 2)
:. R
70 N + 700 N = 770 N
weight W
c) Acceleration downwards. Applying Newton's second law:
mass X acceleration
resultant downwards force (W- R)
(70 kg x 2.0 m s- 2)
700 N - R
:. R
7 00 N - 140 N = 560 N
d) Constant speed implies no acceleration, so resultant force = 0
:. contact force acting upwards = weight acting downwards = mass X g
= 70 kg X
10 N kg- 1 = 700 N
Can you see why you feel heavier as a lift accelerates upwards, and lighter as it accelerates downwards?
Normally you feel a contact force from the floor equal to your weight. As the lift accelerates upwards, you
experience a greater contact force, and so you feel heavier. The opposite is true for downward acceleration.
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Example 8
A car of mass 1200 kg tows a caravan of mass 1800 kg
along a level road. They are linked by a rigid tow-bar.
The car accelerates at 1. 80 m s- 2 , pushed by a motive force
of 5600 N. The resistive force on the car due to friction and
air resistance is 65.0 N. Calculate:
a) the resistive force F on the caravan,
b) the tension Tin the tow-bar.
a = 1.80 m s- 2
F~-1
□□
65 N
□
5600 N
---T--T-
a) The tension in the tow-bar pulls equally on the car and on the
caravan. If we start by looking at the car and caravan together
we can ignore tension, as the tension forces balance out.
The car and the caravan must have the same acceleration as
they are linked by a rigid bar.
Applying Newton's second law (F = ma) to the car and the caravan together:
Force
(5600 N - 65 N - F)
5535 N - F
5535 N - F
•
F • •
mass x acceleration
(1200 kg + 1800 kg) x 1.80 m s- 2
3000 kg X 1.80 m s- 2
5400 kg m s- 2 = 5400 N
5535 N - 5400 N = 135 N (3 s.f .)
b) We can find the tension in the tow-bar by applying Newton's
second law to either the car or the caravan. We will do both here.
Always start by drawing free-body force diagrams for each one :
Method 1 : Forces on car (apply F = ma)
65 N
5600 N - 65 N - T
1200 kg x 1.80 m s- 2
.......,.~....._
5535 N - T
2160 kg m s- 2 = 2160 N
r
:. T - 5535 N - 2160 N = 3375 N = 3.38 x 103 N (3 s.f.)
Method 2: Forces on caravan (apply F = ma)
T - 135 N
1800 kg x 1.80 m s- 2
135 N..------LJ
T - 135 N
3240 kg m s- 2 = 3240 N
:. T - 3240 N + 135 N = 3375 N = 3.38 x 103 N (3 s.f.)
5600 N
□□
Example 9
A battery-powered wheelchair can provide a motive force of 170 N.
The person and wheelchair have a combined weight of 1000 N
and the total resistive force acting is 20 N.
What is the maximum angle 0 of the ramp that will allow the
wheelchair to travel up it at a steady speed?
Steady speed means no acceleration, and so
no resultant force along the slope.
So : total force = total force
up slope
down slope
F - R + W sin0
170 N - 20 N + (1000 N X sin 0)
170 N - 20 N
:. sine = 0.15
1000 N
•
sin- 1 0.15 = 8.6° (2 s.f.)
• •
e -
(-)
weight, W
w
59
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