Resolving a 20N force in direction N 60° E

[slingboi]

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

 

[ehild]

Sorry, you need to show your work first. ehild

 

[slingboi]

north = 20Cos(60) = 10
east = 20Sin(60) = 17.3
What could (c) be, does N 30(degrees) E become the hypotenuse?

 

[grzz]

(a) & (b) are correct.

 

[ehild]

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

 

[slingboi]

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

 

[ehild]

I hope so ...

ehild