Masses Connected Over Pulleys


Masses connected by a string over a pulley system is another extension of Newton’s 2nd and 3rd laws. In these situations, the applied force is caused by the weight of the mass hanging over the pulley.

Consider two objects A and B, with masses of 2kg and 5kg respectively, that are connected by a cable. Object A is hanging over the pulley as shown below:

We can consider a few aspects of this problem:

The total force acting on the system is a result of the weight of object A:

{ W }_{ A }=mg

{ W }_{ A }=2\times 9.8

{ W }_{ A }=19.6\quad N

This weight force, { W }_{ A }=19.6\quad N, is the force that accelerates the whole system. By the whole system we are referring to the total mass of 2+5=7\quad kg. Therefore, the acceleration of the system is:

a=\cfrac { F }{ m }

a=\cfrac { 19.6 }{ 7 }

a=2.8\cfrac { m }{ { s }^{ 2 } }


We can now determine the net force acting on each object:

Object A: F=ma

F=2\times 2.8

F=5.6\quad N down

Object B: F=ma. This force is the tension in the cable:

F=5\times 2.8

F=14\quad N right


When we analyse the net forces acting on each object we need to consider the tension in the cable. Tension in strings and cables acts in both directions according to Newtons 3rd law.

In the example above, the force that acts on object B (14 N right) comes from the tension in the cable – a force of 14 N acting to the right. According to Newtons 3rd law, there is an equal and opposite force of 14 N up acting on object A. This also comes from the tension in the cable. Note that whilst Newtons 3rd law refers to equal and opposite forces, it still applies here as the pulley system changes the direction of the force.

We can calculate the net force, F_{ net } acting on object A:

F_{ net }=19.6-14=5.6\quad N\quad down, which agrees with our value above.


Example 1:

The system below has a 9kg mass on a table connected by a string over a pulley system to a 4kg mass. 

Calculate:

a) The acceleration of the system

b) The net force acting on each object

c) The tension in the string

a) The total force acting on the system is a result of the weight of the 4kg mass:

W=mg

W=4\times 9.8

W=39.2\quad N

This weight force, W=39.2\quad N, is the force that accelerates the whole system. By the whole system we are referring to the total mass of 9+4=13\quad kg. Therefore, the acceleration of the system is:

a=\cfrac { F }{ m }

a=\cfrac { 39.2 }{ 13 }

a=3.02\cfrac { m }{ { s }^{ 2 } }

b) Determine the net force acting on each object:

9kg mass: F=ma

F=9\times 3.02

F=27.18\quad N right

4kg mass: F=ma

F=4\times 3.02

F=12.08\quad N down

c) The tension in the string is equal to the force that accelerates the 9kg mass:

Tension = 27.18 N

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