Question

Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 30° and 120°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Answer 1

Magnitude:

`|R|=180.27756`

Direction:

`\theta=63.69^(\circ \:)`

Explanation:

Calculation of first force:

we are given

`|F_1|=150`

`\theta=30`

now, we can find components

`x=|F_1|cos(\theta)`

`x=150cos(30)`

`x=75√(3)`

`y=|F_1|sin(\theta)`

`y=150sin(30)`

`y=75`

now, we can find force

`F_1=(75√(3),75)`

Calculation of Second force:

we are given

`|F_2|=100`

`\theta=120`

now, we can find components

`x=|F_2|cos(\theta)`

`x=100cos(120)`

`x=-50`

`y=|F_2|sin(\theta)`

`y=100sin(120)`

`y=50√(3)`

now, we can find force

`F_2=(-50,50√(3))`

now, we can find resultant force

`R=F_1+F_2`

`R=(75√(3),75)+(-50,50√(3))`

`R=(75√(3)-50,75+50√(3))`

`R=(79.90381 ,161.60254)`

Magnitude of Resultant:

`|R|=√(79.90381^2+161.60254^2)`

`|R|=180.27756`

Direction of Resultant:

`\theta=tan^(-1)((161.60254)/(79.90381))`

`\theta=63.69^(\circ \:)`

Answer 2

`Magnitude=180.27 \ lbf \\ \\ Direction=63.69 \ degrees`

Explanation:

We have two forces as follows:

First force:

Magnitude: 150 pounds

Angle: 30°

First force:

Magnitude: 100 pounds

Angle: 120°

So the components can be found as follows:

`F_(1x)=150cos(30)=129.90 \ lbf\\F_(1y)=150sin(30) = 75 \ lbf \\ \\ F_(2x)=100cos(120)=-50 \ lbf\\F_(2y)=100sin(120) = 86.60 \ lbf`

So the components of the resultant force can be found by adding each component of the individual forces as follows:

`R_(x)=\Sigma F_(x) \\ R_(y)=\Sigma F_(y) \\ \\ R_(x)=129.90-50=79.90 \ lbf \\ R_(y)=75+86.60=161.6 \ lbf`

Finally, the magnitude and direction of the resultant force is:

`Magnitude \rightarrow R=\sqrt{R_(x)^2+R_(y)^2}=√(79.90^2+161.6^2)=180.27 \ lbf \\ \\ Direction \rightarrow \theta=tan^(-1)((R_(y))/(R_(x)))=tan^(-1)((161.6)/(79.90))=63.69 \ degrees`