Answer to Question #155871 in Physics for muhammad tajammal

Question

A rope is tied to the top of a 10 m flag pole. The rope is tensioned to 120N and then tied to a stake in the ground 20m from the bottom of the pole. The ground is level. Find the vertical and horizontal components of this force.

Answer 1

The situation is shown in the figure. From the right triangle $\triangle ABC$, obtain:

$$AB = \sqrt{CB^2 + AC^2} = \sqrt{10^2 + 20^2} = 10\sqrt{5}m$$

The vertical projection of the force $\mathbf{F}$ is:

$$F_v = F\sin\theta$$

According to the sine definition, obtain:

$$F_v = F\dfrac{CB}{AB} = 120\dfrac{10}{10\sqrt{5}} = \dfrac{120}{\sqrt{5}} = 24\sqrt{5}N$$

The horizontal projection is (according to the cosine definition):

$$F_h = F\cos\theta= F\dfrac{AC}{AB} = 120\dfrac{20}{10\sqrt{5}} = 48\sqrt{5}N$$

Answer. Horizontal $48\sqrt{5}N$, vertical $24\sqrt{5}N$.