A Level Maths: Connected Particles
Introduction
Here we are going to look at some year 13 A Level maths mechanics. The skills that you learned in year 12 are still going to be used.
The questions will obviously be more in depth as you are looking at objects on a slope as well as the concept of friction is also used. You will be resolving forces parallel and perpendicular to the plane and so you need to ensure that you have all your directions correct and that you have a good clear diagram. You will need to be able to do the questions without a diagram so that will be your starting point if a diagram is not part of the question.
When you draw a diagram it is important that you are including all the relevant forces.
Connected Particles
When two objects are connected together we generally take the tension in the string/rope/tow bar to be the same throughout. As well as this we only consider the string/rope/tow bar to be taut in order to ensure that two or more objects have the same acceleration.
The key to dealing with connected particles is to draw a clear diagram and to consider each object separately.
Newton’s Law of F=m a will be used in order to determine the equation of motion.
Example
A train of mass 500 kg is pulled up a slope inclined at 30° to the horizontal. At one end of the train is a rope which passes over a smooth pulley which is connected to a free hanging mass of 1000kg. There is a force of resistance of 100N acting on the train.
What is the acceleration of the train and the tension in the rope?
Solution
Below is a diagram showing the train, the free hanging mass and all forces.
In order to find the tension in the rope and the acceleration we need to consider each object separately.
\begin{aligned} &1000 g-T=1000 a\\ &T-100-500 g \sin 30=500 a \end{aligned}Rearranging the first equation for T gives T=1000 q-1000 a and substituting this into the second equation will give:
\begin{gathered} 1000 g-1000 a-100-500 g \sin 30=500 a \\ 1000 g-100-500 g \sin 30=1500 a \\ a=\frac{1000 g-100-500 g \sin 30}{1500}=4.83 \mathrm{~ms}^{-2} \\ \therefore T=1000 g-1000(4.83)=4970 \mathrm{~N} \end{gathered}Example
A car of mass 900kg pulls a caravan of mass 700kg along a level horizontal road. The car exerts a forward force of 2.4kN and there is no resistance to motion. What is the acceleration of the car and the caravan and the tension in the tow bar?
Solution
Remember it is essential practice in mechanics to draw a diagram as the one above to answer the question.
Considering each object we have:
\begin{gathered} 2400-T-900 a \\ T=700 a \end{gathered}Adding these equations we have:
2400=1600 a \therefore a=1.5 \mathrm{~ms}^{-2} \text { and } T=700(1.5)=1050 \mathrm{~N}So far, two questions have been covered and each has required the use of a diagram. The first question has required the use of resolving forces parallel and perpendicular to the plane. This is an essential skill to be able to do and if you are not sure about this concept remember it is just trigonometry. For some it can be a challenge so a mechanics A Level online maths tutor would be highly suggested.
Example
Example
A car of mass 1100kg pulls a caravan of mass 900kg using a tow bar. The driving force on the car is 3000N. Resistance forces of 300N and 450N act on the car and caravan respectively.
What is the acceleration of the car and caravan and the tension in the tow bar?
Solution
There are two ways in which we can approach this problem and both methods will be looked at.
One is to treat each object individually and the other is to treat the system as a whole. Which method is used is a matter of preference.
Treating each object individually:
\begin{aligned} & 3000-300-T=1100 a \\ & T-450=900 a \end{aligned}Adding we obtain:
\begin{gathered} 2250=2000 a \therefore a=1.125 \mathrm{~ms}^{-2} \\ T=450+900(1.125)=1462.5 \mathrm{~N} \end{gathered}Treating the system as a whole
Total mass = 2000kg
Total resistance = 750N
Total driving force = 3000N
Using F=m a \text { we have: } 3000-750=2000 a \therefore a=1.125 \mathrm{~ms}^{-2}
In order to find the tension in the tow bar we can consider the caravan only:
T-450=900(1.125)=1462.5 \mathrm{~N}Connected particles with friction
An example will be used to show how to solve questions relating to connecting particles and friction.
Example
A 6kg blocks sits on a rough horizontal table. Attached to the block is a light string which passes over a smooth pulley which is connected to a free hanging block of mass 4kg. Given that the coefficient of friction is 0.5 and the system is released from rest, find:
- The magnitude of the frictional force
- The acceleration of the system
- The tension in the string
Solution
As usual a clear detailed diagram should be drawn as the one below.
\text { b) Using } F=m a \text { on each block }\begin{aligned} & T-29.4=6 a \\ & 4 g-T=4 a \end{aligned}
Adding gives us:
4 g-29.4=10 a \therefore a=0.98 \mathrm{~ms}^{-2}Part c) T=29.4+6(0.98)=35.28 N
All of these questions have been done with the aid of a diagram. It cannot be stressed enough that a diagram is needed to do mechanics questions. Over our 4 day Easter revision course for A Level maths, questions like the ones shown here will be covered and this is an area that many students find challenging.
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