CAPÍTULO V. Medidas de Protección Aplicables a Trabajos con Encofrados Horizontales
2. Medidas de Protección Colectiva
2.3. Redes Bajo Forjado
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? Attach a weight to a spring balance and move it up and down. What happens to the pointer on the balance?What would you observe if you stood on some bathroom scales in a moving lift?
Hold a heavy book on your hand and move it up and down.
What force do you feel on your hand?
Equation of motion
Suppose you make the book accelerate upwards at a m s 2. Figure 4.1 shows the forces acting on the book and the acceleration.
Figure 4.1
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the acceleration is shown with a different type of arrow
the weight is always shown as mg for moving bodies forces acting
on the book
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By Newton’s first law, a resultant force is required to produce an acceleration. In this case the resultant upward force is R mg newtons.
You were introduced to Newton’s second law in Chapter 3. When the forces are in newtons, the mass in kilograms and the acceleration in metres per second squared, this law is:
Resultant force mass a
So for the book: R mg ma When Newton’s second law is applied, the resulting equation is called the equation of motion.
When you give a book of mass 0.8 kg an acceleration of 0.5 m s 2 equation becomes
R 0.8 10 0.8 0.5 R 8.4
When the book is accelerating upwards the reaction force of your hand on the book is 8.4 N. This is equal and opposite to the force experienced by you so the book feels heavier than its actual weight, mg, which is 0.8 10 8 N.
EXERCISE 4A 1 Calculate the resultant force in newtons required to produce the following accelerations.
(i) A car of mass 400 kg has acceleration 2 m s 2.
(ii) A blue whale of mass 177 tonnes has acceleration 12 m s–2.
(iii) A pygmy mouse of mass 7.5 g has acceleration 3 m s–2.
(iv) A freight train of mass 42 000 tonnes brakes with deceleration 0.02 m s–2.
(v) A bacterium of mass 2 10–16 g has acceleration 0.4 m s–2.
(vi) A woman of mass 56 kg falling off a high building has acceleration 9.8 m s–2.
(vii) A jumping flea of mass 0.05 mg accelerates at 1750 m s–2 during take-off.
(viii) A galaxy of mass 1042 kg has acceleration 10–12 m s–2.
2 A resultant force of 100 N is applied to a body. Calculate the mass of the body when its acceleration is
Reaction of hand Weight of book, i.e.
mass of book g
Applying Newton’s second law along a line
EXAMPLE 4.1 A lift and its passengers have a total mass of 400 kg. Find the tension in the cable supporting the lift when
Before starting the calculations you must define a direction as positive. In this example the upward direction is chosen to be positive.
(i) At rest
As the lift is at rest the forces must be in equilibrium. The equation of motion is
T mg 0 T 400 10 0
T 4000 The tension in the cable is 4000 N.
(ii) Moving at constant speed
Again, the forces on the lift must be in equilibrium because it is moving at a constant speed, so the tension is 4000 N.
(iii) Accelerating upwards
The resultant upward force on the lift is T – mg so the equation of motion is T – mg = ma
which in this case gives T – 400 10 = 400 0.8 T – 4000 = 320 T = 4320 The tension in the cable is 4320 N.
(iv) Accelerating downwards The equation of motion is
T – mg = ma In this case, a is negative so
T – 400 10 = 400 (–0.6) T – 4000 = –240 T = 3760 The tension in the cable is 3760 N.
Figure 4.2
Newton’s second law
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? How is it possible for the tension to be 3760 N upwards but the lift to acceleratedownwards?
EXAMPLE 4.2 This example shows how the suvat formulae for motion with constant
acceleration, which you met in Chapter 2, can be used with Newton’s second law.
A supertanker of mass 500 000 tonnes is travelling at a speed of 10 m s 1 when its engines fail. It then takes half an hour for the supertanker to stop.
(i) Find the force of resistance, assuming it to be constant, acting on the supertanker.
When the engines have been repaired it takes the supertanker 10 minutes to return to its full speed of 10 m s 1.
(ii) Find the driving force produced by the engines, assuming this also to be constant.
SOLUTION
Use the direction of motion as positive.
(i) First find the acceleration of the supertanker, which is constant for constant forces. Figure 4.3 shows the velocities and acceleration.
Figure 4.3
Since the supertanker is slowing down, you expect a to be negative.
You have to be very careful with signs: the resultant force and acceleration are both positive
towards the right.
Applying Newton’s second law along a line
You know u 10, v 0, t 1800 and you want a, so use v u at.
0 10 1800a a
Now we can use Newton’s second law (Newton II) to write down the equation of motion. Figure 4.4 shows the horizontal forces and the acceleration.
Using Newton’s second law again D R 500 000 000 a D 2.78 106 500 000 000
D 2.78 106 8.33 106
The driving force is 11.11 106 N or 11 100 kN (correct to 3 s.f.).
Tackling mechanics problems
When you tackle mechanics problems such as these you will find them easier if you:
L always draw a clear diagram
L clearly indicate the positive direction
L label each object (A, B, etc. or whatever is appropriate)
L show all the forces acting on each object
L make it clear which object you are referring to when writing an equation of motion.
The acceleration is negative because the supertanker is slowing down.
Figure 4.4
The upthrust of the water balances the weight of the supertanker in
the vertical direction.
EXERCISE 4B 1 A man pushes a car of mass 400 kg on level ground with a force of 200 N. The car is initially at rest and the man maintains this force until the car reaches a speed of 5 m s 1. Ignoring any resistance forces, find
(i) the acceleration of the car
(ii) the distance the car travels while the man is pushing.
2 The engine of a car of mass 1.2 tonnes can produce a driving force of 2000 N.
Ignoring any resistance forces, find
(i) the car’s resulting acceleration
(ii) the time taken for the car to go from rest to 27 m s 1 (about 60 mph).
3 A top sprinter of mass 65 kg starting from rest reaches a speed of 10 m s 1 in 2 s.
(i) Calculate the force required to produce this acceleration, assuming it is uniform.
(ii) Compare this to the force exerted by a weight lifter holding a mass of 180 kg above the ground.
4 An ice skater of mass 65 kg is initially moving with speed 2 m s 1 and glides to a halt over a distance of 10 m. Assuming that the force of resistance is constant, find
(i) the size of the resistance force
(ii) the distance he would travel gliding to rest from an initial speed of 6 m s 1
(iii) the force he would need to apply to maintain a steady speed of 10 m s 1.
5 A helicopter of mass 1000 kg is taking off vertically.
(i) Draw a labelled diagram showing the forces on the helicopter as it lifts off and the direction of its acceleration.
(ii) Its initial upward acceleration is 1.5 m s 2. Calculate the upward force its rotors exert. Ignore the effects of air resistance.
6 Pat and Nicholas are controlling the movement of a canal barge by means of long ropes attached to each end. The tension in the ropes may be assumed to be horizontal and parallel to the line and direction of motion of the barge, as shown in the diagrams.
The mass of the barge is 12 tonnes and the total resistance to forward motion may be taken to be 250 N at all times. Initially Pat pulls the barge forwards from rest with a force of 400 N and Nicholas leaves his rope slack.
Exercise 4B
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Applying Newton’s second law along a line
Pat continues to pull with the same force until the barge has moved 10 m.
(ii) What is the speed of the barge at this time and for what length of time did Pat pull?
Pat now lets her rope go slack and Nicholas brings the barge to rest by pulling with a constant force of 150 N.
(iii) Calculate
(a) how long it takes the barge to come to rest
(b) the total distance travelled by the barge from when it first moved
(c) the total time taken for the motion. [MEI]
7 A spaceship of mass 5000 kg is stationary in deep space. It fires its engines, producing a forward thrust of 2000 N for 2.5 minutes, and then turns them off.
(i) What is the speed of the spaceship at the end of the 2.5 minute period?
(ii) Describe the subsequent motion of the spaceship.
The spaceship then enters a cloud of interstellar dust which brings it to a halt after a further distance of 7200 km.
(iii) What is the force of resistance (assumed constant) on the spaceship from the interstellar dust cloud?
The spaceship is travelling in convoy with another spaceship which is the same in all respects except that it is carrying an extra 500 kg of equipment. The second spaceship carries out exactly the same procedure as the first one.
(iv) Which spaceship travels further into the dust cloud?
8 A crane is used to lift a hopper full of cement to a height of 20 m on a building site. The hopper has mass 200 kg and the cement 500 kg. Initially the hopper accelerates upwards at 0.05 m s 2, then it travels at constant speed for some time before decelerating at 0.1 m s 2 until it is at rest. The hopper is then emptied.
(i) Find the tension in the crane’s cable during each of the three phases of the motion and after emptying.
The cable’s maximum safe load is 10 000 N.
(ii) What is the greatest mass of cement that can safely be transported in the same manner?
The cable is in fact faulty and on a later occasion breaks without the hopper leaving the ground. On that occasion the hopper is loaded with 720 kg of cement.
(iii) What can you say about the strength of the cable?
9 The police estimate that for good road conditions the frictional force, F, on a skidding vehicle of mass m is given by F = 0.8 mg. A car of mass 450 kg skids to a halt narrowly missing a child. The police measure the skid marks and find they are 12.0 m long.
(i) Calculate the deceleration of the car when it was skidding to a halt.
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