Please can you help me to resolve this vector force?

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The discussion focuses on resolving a 20N force directed N 60° E into its components. The resolved parts are calculated as 10N north and 17.3N east using trigonometric functions. Participants clarify the need to find the projection of the force in the direction of N 30° E, identifying the angle between the force and this direction as 30 degrees. The calculation for this projection involves using the cosine of 30 degrees. The conversation emphasizes the importance of showing work in solving vector problems.
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A force P is 20N in a direction N 60(degrees) E. What is the resolved part of the force
(a) north
(b) east
(c) N 30(degrees) E

I have an idea of what to do for a and b but can you please show me your working aswell, thanks
 
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Sorry, you need to show your work first. ehild
 
north = 20Cos(60) = 10
east = 20Sin(60) = 17.3
What could (c) be, does N 30(degrees) E become the hypotenuse?
 
(a) & (b) are correct.
 
slingboi said:
north = 20Cos(60) = 10
east = 20Sin(60) = 17.3
What could (c) be, does N 30(degrees) E become the hypotenuse?

See attached figure. You need the projection of the force on the direction N 30° E. What is the angle between the force and the given direction?

ehild
 

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30 degrees, oh I see now so would the force N 30(degrees) E be 20cos(30)?
 
I hope so ...

ehild
 
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