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.

The distance, s, 11.1 m. Find F.

Answer 1

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
`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=(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
`T` 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`