I need some help with a vectors question.
Two forces f1 and f2 act on a particle:
f1 = -3i+7j newtons
f2= i-j newtons
The resultant force R acting on the particle is given by R=f1+f2
a) Calculate the magnitude of R in Newtons
b)Calculate to the nearest degree, the angle between the line of action R and the vector j.
I have done part a) R=2√10 but i have no clue how to go about part b)
Any help would be greatly appreciated, thank you.
f1 = -3i+7j newtons
f2= i-j newtons
The resultant force R acting on the particle is given by R=f1+f2
a) Calculate the magnitude of R in Newtons
b)Calculate to the nearest degree, the angle between the line of action R and the vector j.
I have done part a) R=2√10 but i have no clue how to go about part b)
Any help would be greatly appreciated, thank you.
Original post by Queenofjazz04
Two forces f1 and f2 act on a particle:
f1 = -3i+7j newtons
f2= i-j newtons
The resultant force R acting on the particle is given by R=f1+f2
a) Calculate the magnitude of R in Newtons
b)Calculate to the nearest degree, the angle between the line of action R and the vector j.
I have done part a) R=2√10 but i have no clue how to go about part b)
Any help would be greatly appreciated, thank you.
f1 = -3i+7j newtons
f2= i-j newtons
The resultant force R acting on the particle is given by R=f1+f2
a) Calculate the magnitude of R in Newtons
b)Calculate to the nearest degree, the angle between the line of action R and the vector j.
I have done part a) R=2√10 but i have no clue how to go about part b)
Any help would be greatly appreciated, thank you.
Sketch R and a simple bit of trig.
For any vector, the angle with the x axis is arctan(y/x) where y and x are the components of the vector.
This is pretty much as standard as the magnitude of ai+bj=sqrt (a^2+b^2). In this case the angle with the x axis is arctan (b/a).
Because you are asked for the angle with j then you need to subtract from 90.
Don't forget than arctan will always give you an angle between -90 and +90 which in this case is particularly useful.
as @mbq2766 says, always sketch a diagram.
This is pretty much as standard as the magnitude of ai+bj=sqrt (a^2+b^2). In this case the angle with the x axis is arctan (b/a).
Because you are asked for the angle with j then you need to subtract from 90.
Don't forget than arctan will always give you an angle between -90 and +90 which in this case is particularly useful.
as @mbq2766 says, always sketch a diagram.
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