Question

A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.

Answer 1

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

$\sum{F_{x}}=-Tcos\theta + F=0$ (1)

$\sum{F_{y}}=Tsin\theta - mg=0$ (2)

Where $T$ is the tension force of the rope, $m=28 kg$ the mass, $g=9.8 m/s^{2}$ the acceleration due gravity and $mg$ is the weight.

On the other hand, we can calculate $\theta$ as follows:

$cos\theta=\frac{s}{l}$

$\theta=cos^{-1}(\frac{s}{l})$

Where $s=11.1 m$ and $l=18 m$

$\theta=cos^{-1}(\frac{11.1 m}{18 m})$

$\theta=51.9\°$ (3)

Now, we firstly need to find $T$ from (2):

$T=\frac{mg}{sin\theta}$ (4)

$T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}$

$T=348.69 N$ (5)

Substituting (5) in (1):

$F=Tcos\theta$ (6)

$F=348.69 N cos(51.9\°)$

Finally:

$F=215.15 N$