To make a stationary object move or to change the direction in which the object is moving requires a force to be applied externally to the object. This concept is known asNewton’s first law of motion and may be stated as:
An object remains in a state of rest, or continues in a state of uniform motion in a straight line, unless it is acted on by an externally applied force
Since a force is necessary to produce a change of motion, an object must have some resistance to a change in its motion. The force necessary to give a stationary pram a given acceleration is far less than the force necessary to give a stationary car the same acceleration on the same surface. The resistance to a change in motion is called theinertiaof an object and the amount of inertia depends on the mass of the object. Since a car has a much larger mass than a pram, the inertia of a car is much larger than that of a pram.
Newton’s second law of motionmay be stated as: The acceleration of an object acted upon by an external force is proportional to the force and is in the same direction as the force
Thus, forceα acceleration, or force=a constant× acceleration, this constant of proportionality being the mass of the object, i.e.
force=mass×acceleration
The unit of force is the newton (N) and is defined in terms of mass and acceleration. One newton is the force required to give a mass of 1 kilogram an acceleration of 1 metre per second squared. Thus
F =ma
whereF is the force in newtons (N),mis the mass in kilograms (kg) andais the acceleration in metres per second squared (m/s2), i.e. 1 N= 1 kg m
s2
It follows that 1 m/s2 = 1 N/kg. Hence a gravi- tational acceleration of 9.8 m/s2 is the same as a gravitational field of 9.8 N/kg.
Newton’s third law of motionmay be stated as: For every force, there is an equal and opposite reacting force
Thus, an object on, say, a table, exerts a downward force on the table and the table exerts an equal upward force on the object, known as a reaction forceor just areaction.
Problem 1. Calculate the force needed to accelerate a boat of mass 20 tonne uniformly from rest to a speed of 21.6 km/h in
10 minutes.
The mass of the boat,m, is 20 t, that is 20000 kg. The law of motion, v = u +at can be used to determine the accelerationa.
The initial velocity,u, is zero,
the final velocity,v =21.6 km/h= 21.6
3.6 =6 m/s, and the time,t =10 min=600 s.
Thusv =u+at, i.e. 6=0+a×600, from which,
a= 6
600 =0.01 m/s 2
From Newton’s second law,F =ma i.e. force=20000×0.01 N=200 N
Problem 2. The moving head of a machine tool requires a force of 1.2 N to bring it to
rest in 0.8 s from a cutting speed of
30 m/min. Find the mass of the moving head.
From Newton’s second law,F =ma, thusm= F a, where force is given as 1.2 N. The law of motion v=u+at can be used to find accelerationa, where v = 0, u = 30 m/min = 30 60 m/s = 0.5 m/s, and t =0.8 s. Thus, 0=0.5+a×0.8 from which, a= −0.5 0.8 = −0.625 m/s 2 or a retardation of 0.625 m/s2. Thus themass,m= F
a = 1.2
0.625 =1.92 kg
Problem 3. A lorry of mass 1350 kg accelerates uniformly from 9 km/h to reach a velocity of 45 km/h in 18 s. Determine (a) the acceleration of the lorry, (b) the uniform force needed to accelerate the lorry.
(a) The law of motion v =u+at can be used to determine the acceleration, where final velocity v= 45 3.6 m/s, initial velocityu= 9 3.6 m/s and timet =18 s. Thus 45 3.6 = 9 3.6+a×18 from which, a= 1 18 45 3.6− 9 3.6 = 1 18 36 3.6 = 10 18 = 5 9 m/s 2 or0.556 m/s2
(b) From Newton’s second law of motion, force,F =ma=1350×5
9 =750 N
Problem 4. Find the weight of an object of mass 1.6 kg at a point on the earth’s surface where the gravitational field is 9.81 N/kg (or 9.81 m/s2).
The weight of an object is the force acting vertically downwards due to the force of gravity acting on the object. Thus:
weight=force acting vertically downwards =mass×gravitational field
=1.6×9.81=15.696 N
Problem 5. A bucket of cement of mass 40 kg is tied to the end of a rope connected to a hoist. Calculate the tension in the rope when the bucket is suspended but stationary. Take the gravitational field, g, as 9.81 N/kg (or 9.81 m/s2).
The tension in the rope is the same as the force acting in the rope. The force acting vertically down- wards due to the weight of the bucket must be equal to the force acting upwards in the rope, i.e. the tension
Weight of bucket of cement,
F =mg=40×9.81=392.4 N Thus,the tension in the rope = 392.4 N
Problem 6. The bucket of cement in Problem 5 is now hoisted vertically upwards with a uniform acceleration of 0.4 m/s2. Calculate the tension in the rope during the period of acceleration.
With reference to Figure 13.1, the forces acting on the bucket are:
(i) a tension (or force) ofT acting in the rope (ii) a force ofmg acting vertically downwards, i.e.
the weight of the bucket and cement
Acceleration Force due to acceleration F= ma Weight, mg T Figure 13.1
The resultant forceF =T −mg;
hence, ma=T −mg
i.e. 40×0.4=T −40×9.81 from which, tension,T =408.4 N
By comparing this result with that of Problem 5, it can be seen that there is an increase in the tension in the rope when an object is accelerating upwards.
Problem 7. The bucket of cement in Problem 5 is now lowered vertically downwards with a uniform acceleration of 1.4 m/s2. Calculate the tension in the rope during the period of acceleration.
With reference to Figure 13.2, the forces acting on the bucket are:
(i) a tension (or force) of T acting vertically upwards
(ii) a force ofmgacting vertically downwards, i.e. the weight of the bucket and cement
Acceleration
Weight, mg
T F= ma
Figure 13.2
The resultant force, F =mg−T
Hence, ma=mg−T
from which,tension, T =m(g−a)
=40(9.81−1.4)
=336.4 N
By comparing this result with that of Problem 5, it can be seen that there is a decrease in the tension in the rope when an object is accelerating down- wards.
Now try the following exercise
Exercise 63 Further problems on New-