NLM O6

Solution

We analyze the equilibrium of a pulley on a string to determine the tension.

Since the pulley is frictionless, the tension $T$ is the same throughout the string. For horizontal equilibrium of the pulley, the angles the two string segments make with the vertical must be equal ($\theta$).

Let the lengths of the two string segments be $L_1$ and $L_2$. Total length $L = L_1 + L_2 = 15$ m.
The horizontal projection of the string segments must sum to the distance between the walls ($D = 9$ m).

$$ L_1 \sin \theta + L_2 \sin \theta = D $$ $$ (L_1 + L_2) \sin \theta = D $$ $$ 15 \sin \theta = 9 $$ $$ \sin \theta = \frac{9}{15} = \frac{3}{5} $$

This result holds regardless of the vertical level difference between the pegs. The angle $\theta$ with the vertical is determined solely by the ratio of horizontal separation to string length.

Now consider vertical equilibrium. The upward components of tension balance the total load ($W = W_p + W_b = 4.8 + 8.0 = 12.8$ N).

$$ 2T \cos \theta = W $$

Since $\sin \theta = 3/5$, we have $\cos \theta = 4/5$.

$$ 2T \left( \frac{4}{5} \right) = 12.8 $$ $$ \frac{8T}{5} = 12.8 $$ $$ T = \frac{12.8 \times 5}{8} = 1.6 \times 5 = 8.0 \text{ N} $$

The tension is 8.0 N. Since the calculation of $\theta$ (and thus $\cos\theta$) did not depend on the vertical positions of the pegs, this tension value is independent of the level difference.