Question

Can someone please help me ​

Answer 1

F = 153 N

`\theta = 11.3°`

Step-by-step explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components
`F_(x)` and
`F_(y)`:

`F_(x)` = 350 N - 200 N = 150 N

`F_(y)` = 180 N - 150 N = 30 N

The magnitude of the resultant force F is given by

`F = \sqrt{F_(x)^(2)+F_(y)^(2)}`

`\:\:\:\:\:\:= \sqrt{(150\:N)^(2)+(30\:N)^(2)}`

`\:\:\:\:\:\:=153\:N`

To find the direction
`\theta`, we use

`\tan \theta = (F_(y))/(F_(x))=(30\:N)/(150\:N)=0.2`

or

`\theta = \tan^(-1)(0.2) = 11.3°`