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. The distance, s, 11.1 m. Find F.

Question

asked May 18, 2020 94.1k views

1 Answer

Steve Vinoski · Answer 1 · 2020-05-23T03:52:25+0000

Answer: 215.15 N

Step-by-step 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
is the tension force of the rope,
m=28 kg the mass,
g=9.8 m/s^(2) the acceleration due gravity and
is the weight.

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

$cos\theta=(s)/(l)$

$\theta=cos^(-1)((s)/(l))$

Where
s=11.1 m and
l=18 m

$\theta=cos^(-1)((11.1 m)/(18 m))$

$\theta=51.9\°$ (3)

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

$T=(mg)/(sin\theta)$ (4)

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

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