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