Teach A Level Maths Connected Particles (2). Volume 4: Mechanics 1 Connected Particles (2) Volume 4: Mechanics 1 Connected Particles (2)

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Teach A Level Maths Connected Particles (2)

Volume 4: Mechanics 1 Connected Particles (2) Volume 4: Mechanics 1 Connected Particles (2)

car v a R FdFd The force due to the car’s engine is called the driving force. F d Mg Suppose a car is moving along a straight horizontal road. Modelling the car as a particle we have the following force diagram. Fr 1 We may also have a force resisting the motion. Suppose the car is towing a trailer using a towrope to connect the two.

car v R FdFd Mg Fr 1 a Tip: Exaggerate the length of the tow rope so that there is room to show the forces. The tension in the towrope is the force that accelerates the trailer. T mg N T Fr 2

To find an equation for T, we need to use F=ma on either the car or the trailer. car v R FdFd Mg Fr 1 a T mg N T Fr 2 We draw exactly the same force diagram for a towbar or coupling but the force in the bar may be a thrust instead of a tension. If the force is a thrust, T will be negative.

car v R FdFd Mg Fr 1 a T mg N T Fr 2 You can find the driving force, F d if you are given the values of the acceleration and the resisting forces using F = ma. You must consider the forces on each mass in turn.

e.g.1.A car of mass 1000 kg is towing a trailer of mass 200 kg in a straight horizontal line. The car is travelling with constant acceleration of magnitude 0·5 m s -2. The resistance to motion for the car is of magnitude 600 newtons and the trailer is 100 newtons. The towrope connecting the car and trailer is horizontal. Find the magnitudes of (a) the tension in the towrope and (b) the driving force.

1000g R FdFd v T 600 car trailer car: mass 1000 kg resistance: 600 newtons

1000g 200g 100 N R FdFd v T T 600 car trailer trailer: mass 200 kg resistance: 100 newtons We always show all the forces, even though the normal reactions are not used in this type of problem.

1000g 200g acceleration: 0·5 m s -2 100 N R FdFd v T T 600 car trailer a = 0·5 (a) Find T Trailer: N2L: Resultant force  mass  acceleration T100  200  T  200  0·5

1000g 200g acceleration: 0·5 m s -2 100 N R FdFd v T T 600 car trailer Trailer: N2L: Resultant force  mass  acceleration T100  200  T  200  0·5 Consider Car alone: FdFd  T 600   1000  0·5 Subs. for T : F d  300 The tension in the towrope is 200 newtons and the driving force is 1300 newtons. a = 0·5 (b) Find F d (a) Find T

 We solve problems by using N2L on the bodies separately and/or together. SUMMARY  When a vehicle tows another with a rope, there is a tension in the rope.  The tension ( or thrust ) provides the force that gives the towed vehicle an acceleration equal to that of the towing vehicle.  If a towbar or coupling is used, we draw the same diagram. If the calculations give a negative value of T there is a thrust instead of a tension.

EXERCISE 1.The sketch shows a car towing a caravan. The masses and resistances to motion are shown. (a) The driving force of the car is 1700 newtons (b)If the car then decelerates at a constant rate of 0·3 m s -2, find (i)the force in the tow bar, explaining the sign in your answer, and (ii)the new driving force. 1500 kg 500 kg 800 N 100 N Draw a complete force diagram and show that the acceleration of the car and caravan is 0·4 m s -2.

car caravan 500g R 1700 T 100 1500g 800 N T driving force: 1700 newtons (a) Find a. 1500 kg 500 kg 800 N 100 N a Car N2L: Resultant force  mass  acceleration 1700800  1500a  T  a  800 2000  a  0·4 m s -2 Solution: EXERCISE caravan: 100  500a T adding 1700-800-100 = 2000a

500g R 1700 T 100 car caravan 1500g 800 N T (b)(i) Find T a 1 =  0·3 caravan: N2L: Resultant force  mass  acceleration FdFd 100  500(  0·3) T  T   50    150 + 100 T The negative sign shows that the force is the opposite direction to that shown in the diagram. ( We say there is a thrust of 50 newtons. ) Solution: EXERCISE

500g R 1700 T 100 car caravan 1500g 800 N T (b)(ii) Find F d a 1 =  0·3 car: N2L: Resultant force  mass  acceleration FdFd  F d  300    450  800  50 FdFd Solution: T   50 800  1500(  0·3) FdFd  T The driving force is 300 newtons EXERCISE

The following page contains the summary in a form suitable for photocopying.

Summary CONNECTED PARTICLES (2) TEACH A LEVEL MATHS – MECHANICS 1  We solve problems by using N2L on the bodies separately and/or together.  When a vehicle tows another with a rope, there is a tension in the rope.  The tension ( or thrust ) provides the force that gives the towed vehicle an acceleration equal to that of the towing vehicle.  If a towbar or coupling is used, we draw the same diagram. If the calculations give a negative value of T there is a thrust instead of a tension.

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