Question

Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent

is 12/5.​

Answer 1

`F_r = 200N`

Step-by-step explanation:

Given

Let the two forces be

`F_1 = 130N`

`F_2 = 110N`

and

`\tan(\theta) = (12)/(5)`

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

`F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)`

This means that we need to first calculate
`\cos(\theta)`

Given that:

`\tan(\theta) = (12)/(5)`

In trigonometry:

`\tan(\theta) = (Opposite)/(Adjacent)`

By comparing the above formula to
`\tan(\theta) = (12)/(5)`

`Opposite = 12`

`Adjacent = 5`

The hypotenuse is calculated as thus:

`Hypotenuse^2 = Opposite^2 + Adjacent^2`

`Hypotenuse^2 = 12^2 + 5^2`

`Hypotenuse^2 = 144 + 25`

`Hypotenuse^2 = 169`

`Hypotenuse = \sqrt{169`

`Hypotenuse = 13`

`\cos(\theta)` is then calculated using:

`\cos(\theta)= (Adjacent)/(Hypotenuse)`

`\cos(\theta)= (5)/(13)`

Substitute values for
`F_1`,
`F_2` and
`cos(\theta)` in

`F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)`

`F_r^2 = 130^2 + 110^2 + 2*130*110*(5)/(13)`

`F_r^2 = 16900 + 12100 + 11000`

`F_r^2 = 40000`

Take square roots of both sides

`F_r = \sqrt{40000`

`F_r = 200N`

Hence, the resultant force is 200N