Masses Connected by Vertical Strings - Learn

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: