Question

A physical therapist gone wild has constructed the (stationary) assembly of massless pulleys and cords seen in Fig. 12-24. One long cord wraps around all the pulleys, and shorter cords suspend pulleys from the ceiling or weights from the pulleys. Except for one, the weights (in newtons) are indicated.

(a) What is that last weight? (Hint:When a cord loops halfway around a pulley as here, it pulls on the pulley with a net force that is twice the tension in the cord.)

(b) What is the tension in the short cord labeled T?

Short answer

1. The unknown (last) weight is 20 N.
2. Tension in the short cord labeled is 25 N.

Step-by-step solution

The given data

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

Understanding the concept of the pulleys

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 the pulleys.

Formulae:

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

a) Calculation of the unknown weight

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 as it is wound continuously on the pulleys.

Let it be denoted as T'. Let the tension in the green cords be denoted as $F_1,F_2,F_3,F_4$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 23 N hanging from it. The forces acting on this system are T' in each of the red ropes and $F_1$in the green cord. All three are directed upwards, whereas the weight is directed downwards.

Now the force equation for this pulley using the condition of equation (i) can be given as:

$$\begin{gathered} F_1+T\text{'}+T\text{'}=0 \\ {F}_1+2T\text{'}=23..............\left(1\right) \end{gathered}$$

Similarly, for the next pulleys, we write the force equations as:

$$\begin{gathered} F_2-2T\text{'}=15..........\left(2\right) \\ 2T\text{'}=20...........\left(3\right) \\ {F}_3-2T\text{'}=6..........\left(4\right) \\ {F}_4+2T\text{'}=34..........\left(5\right) \\ T-2T\text{'}=5..........\left(6\right) \\ 2T\text{'}=w..............\left(7\right) \end{gathered}$$

Comparing the equations (iii) and (vii), we get the value of the unknown weight as follows:

w = 20 N

Hence, the required weight value is 20 N.

b) Calculation of the tension in the short cord

From part (a) force equations, we can get the value of the tension in the short cord as follows:

We know that,

$2T\text{'}=20N(fromequation(7\left)\right)$

Substituting this value in equation (vi), we get the tension value as:

$$\begin{gathered} T-(20N)=5N \\ T=25N \end{gathered}$$

Hence, the tension in the shorter cord is 25 N.