91Ó°ÊÓ

The pulley system is in equilibrium, having various weights attached to it.

We use the concept of balanced forces to determine the magnitudes of the various tensions and forces acting on the cords. We make use of free-body diagrams for pulleys.

Formulae:

The value of the net force at equilibrium, $F_{net}=0$ (i)

We will consider a free body diagram for each of the pulleys and balance the forces for each.

The tension in each part of the red rope is the same; this is as it wound continuously on the pulleys.

Let it be denoted as T'. Let, the tension in the green cords be denoted as$F_1,F_2,F_3$ , and T as indicated in the diagram. Let the unknown weight be w.

We now consider the first pulley on the right side, to have weight 17 N hanging from it. The forces acting on this system are in each of the red ropes$T\text{'}$and$F_1$in the green cord.

All three are directed upwards, whereas the weight is directed downwards.

Now, the net force equation for this pulley considering equation (i) can be given as:

$$\begin{gathered} F_1+T\text{'}-17=0 \\ {F}_1+2T\text{'}=17.................\left(i\right) \end{gathered}$$

Similarly, for the next pulleys, we write the force equations using equation (i) as follows:

role="math" localid="1660977557551" $$\begin{gathered} 2T\text{'}=10...........\left(ii\right) \\ {F}_3-2T\text{'}=0............\left(iii\right) \\ T-2T\text{'}=0.............\left(iv\right) \\ 3T\text{'}-w=0.............\left(v\right) \end{gathered}$$

From equation (ii), we get,

$T\text{'}=5\mathrm{N}$

And using this value of T’ in equation (v), we find,

$w=15N$

Hence, the value of the weight of the piñata is 15 N .

Also, from equation (iv), we calculate the value of the tension using the value of

$T\text{'}=5N$ can be calculated as:

$$\begin{gathered} T2\left(5\mathrm{N}\right)=0 \\ T=10\mathrm{N} \end{gathered}$$

Hence, the value of the tension of the short cord is 10 N .