Masses Connected by Vertical Strings


Masses connected by vertical strings is another problem that requires both Newton’s 2nd and 3rd laws. Consider two objects A and B, with masses of 2kg and 5kg respectively as shown below.

Let’s examine two situations:

a) Objects A and B are stationary

b) Objects are accelerating up at 2\cfrac { m }{ { s }^{ 2 } }

a) If the objects are stationary, the net force on each object = 0. The forces involved will be the tension in the string. The tension in each string is equal to the weight that they support:

T_{ 1 } = the weight of object A: 

W=mg

W=2\times 9.8

W=19.6\quad N 

T_{ 2 } = the weight of object A + object B:

W=mg. This is due to the string at T_{ 2 } supporting object A and B.

W=7\times 9.8

W=68.6\quad N 

Therefore; T_{ 1 }=19.6N and T_{ 2 }=68.6N


b) If the objects are accelerating up at 2\cfrac { m }{ { s }^{ 2 } } , we can determine the net force acting on each object using F=ma. We can also determine the tension in each string using: T=W-ma. This equation involves vectors and down is best noted as positive. The acceleration will be negative if directed upwards or negative if directed downwards. Note that a positive or negative result for tension is not important because tension is a force that acts in both directions on a string. Essentially this means that the equation: T=W-ma gives us a magnitude for the tension.

If we consider the acceleration of each object we can determine the net force, F_{ net } acting on each object:

F_{ net }=ma

F_{ net }(A)=2\times 2=4N

F_{ net }(B)=5\times 2=10N

Observing the diagram below, we can see that: (we will note down as positive)

T_{ 1 }=W-ma

T_{ 1 }=(2\times 9.8)-(2\times -2)

T_{ 1 }=(19.6)-(-4)

T_{ 1 }=23.6N\quad

T_{ 2 }=W-ma

T_{ 2 }=((2+5)\times 9.8)-((2+5)\times -2)

T_{ 2 }=(7\times 9.8)-(7\times -2)

T_{ 2 }=(68.6)-(-14)

T_{ 2 }=82.6N


Example 1:

Two objects of masses 3kg and 10kg are connected by vertical strings as shown below. The objects are stationary. Calculate:

a) The net force acting on each mass

b) The tension in each string

a) If the objects are stationary, the net force on each object = 0

b) The tension in each string is equal to the weight that they support:

T_{ 1 } = the weight of the 3kg mass: 

W=mg

W=3\times 9.8

W=29.4\quad N 

T_{ 2 } = the weight of both masses:

W=mg

W=13\times 9.8

W=127.4\quad N 

Therefore; T_{ 1 }=29.4N and T_{ 2 }=127.4N


Example 2:

Two objects of masses 5kg and 12kg are connected by vertical cables as shown below. The objects are accelerating upwards at 1.5\cfrac { m }{ { s }^{ 2 } } . Calculate:

a) The tension in each string:

we will note down as positive:

T_{ 1 }=W-ma

T_{ 1 }=(5\times 9.8)-(5\times -1.5)

T_{ 1 }=(49)-(-7.5)

T_{ 1 }=56.5N\quad

T_{ 2 }=W-ma

T_{ 2 }=((5+12)\times 9.8)-((5+12)\times -1.5)

T_{ 2 }=(17\times 9.8)-(17\times -1.5)

T_{ 2 }=(166.6)-(-25.5)

T_{ 2 }=191.5N


Example 3:

Two objects of masses 2kg and 4.5kg are connected by vertical cables as shown below. The objects are accelerating downwards at 3\cfrac { m }{ { s }^{ 2 } } . Calculate:

a) The tension in each string:

we will note down as positive:

T_{ 1 }=W-ma

T_{ 1 }=(2\times 9.8)-(2\times 3)

T_{ 1 }=(19.6)-(6)

T_{ 1 }=13.6N\quad

T_{ 2 }=W-ma

T_{ 2 }=((2+4.5)\times 9.8)-((2+4.5)\times 3)

T_{ 2 }=(6.5\times 9.8)-(6.5\times 3)

T_{ 2 }=(63.7)-(19.5)

T_{ 2 }=44.2N

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