Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 40° and 170° respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

magnitude= 180.278 pounds
((75 3/-50)^2+ (75+50 3)^2)
352500
50 13

direction= 63.90
tan^-1 (( 75+50 3)/(75 3-50))
where you see the big spaces I mage put the check mark symbol

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lublana lublana · Answer 2 · 2019-08-14T08:07:49+02:00

Answer:

Resultant force =115.03 pounds

Direction of resultant force=81.8 degrees

Step-by-step explanation:

We are given that two forces

Let [tex]F_1=150 pounds[/tex]

[tex]F_2=100 pounds[/tex]

[tex]\theta_1=40^{\circ}[/tex]

[tex]\theta_2=170^{\circ}[/tex]

[tex]Sin(40^{\circ})=\frac{vertical component }{hypotenuse }[/tex]

[tex]0.643=\frac{vertcal component }{150}[/tex]

Vertical component of force[tex]F_0.643\times 150=96.45 [/tex]

Horizontal component =[tex]100cos (40^{\circ})=150\times0.766=114.9[/tex]

Vertical component of second force [tex]F_2= 100\times sin170=100\times 0.174=17.4[/tex]

Horizontal component of force [tex]F_2=100cos 170=100\times (-0.985)=-98.5[/tex]

Vertical component of resultant force=96.45+17.4=113.85

Horizontal component of resultant force=114.9-98.5=16.4

Magnitude of resultant force =[tex]\sqrt{(113.85)^2+(16.4)^2}=\sqrt{12961.8225+268.96}=\sqrt{13230.7825}[/tex]=115.03 pounds

Direction =[tex]tan^{-1}(\frac{vertical component}{horizontal component})[/tex]

Direction of resultant vector force=[tex]tan^{-1}(\frac{113.85}{16.4}=tan^{-1}(6.94)[/tex]=81.8 degrees