Help can you please explain this to me thanks (it is all about vector )

AI Thread Summary
The discussion revolves around understanding how to resolve vectors into their x and y components using sine and cosine functions. The key point is that sine is used for the y-component and cosine for the x-component when dealing with angles relative to the horizontal axis. Participants express confusion about when to apply each function, particularly in the context of forces acting at various angles. It is clarified that the angle made with the horizontal determines which function to use: cosine for the adjacent side (x-component) and sine for the opposite side (y-component). Ultimately, understanding the relationship between the angle and the axes is crucial for accurately resolving vector components.
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this problem was discussed by our professor . but I am a lil bit confused on it ! thanks ! help !

FIVE COPLANAR FORCES ACT ON AN OBJECT AND THEIR RESULTANT !

our prof use this formula !

Bx=Bsin(degree)
By=Bcos(degree)

ANSER:

VECTORS X-COMPONENTS Y-COMPONENTS
A 19.0 0
B 7.5 12.99
C -11.31 11.31
D -9.53 -5.5
E 0 -22.0
Σx=5.66 Σy=-3.2

MY QUESTION IS : HOW TO KNOW IF WHAT PARTICULAR OPERATION TO BE USE ON EACH OF THE COMPONENTS ? WHEN WILL I USE THE SIN FUNCTION AND COSINE FUNCTION IN GETTING THE COMPONENTS? THANKS !
 

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Have you studied trigonometry yet?
 
SteamKing said:
Have you studied trigonometry yet?

yes. a year a go. i know how to use the function but my problem is i don't know when to use the cos and sin function like in the x-component my prof use the sin function in computing the x component except on letter D and C . can you please help me how to analyze it . I'm confuse !
 
sanoy87 said:
its easy ,don't worry ,if u r considering component of force & angle between forces & the plane along which component of force under consideration acts - use cosine other wise use sine ,to be exact

can you please elaborate it ? i can't understand :((
 
When in doubt just remember, the side opposite to the angle will have sine and the side adjacent will be the cosine.
 
rock.freak667 said:
when in doubt just remember, the side opposite to the angle will have sine and the side adjacent will be the cosine.

more more :(( please ! I am confuse !
 
Okay,

Lets put it differently. We need to find the resultant force on the body. In fact we need the sum of the forces. As forces are vectors, their direction also needs to be considered in the addition.

Had all the forces been in the same direction, it would have been easier as it would simply be the sum of those.

So what we do is, we choose two directions, preferably perpendicular, which by convention is chosen as the x-axis and y-axis.

We resolve all these forces into these axes. Are you with me?

How we do is like this... if F is the force then F.cos(t) will be the component in x-axis and F.sin(t) in the y axis, if t is the angle made by the force with x-axis.

So we resolve all the forces like this. Sum up all the Fx's (aka force in x-direction) and Fy's

then the resultant force will be ((sum of all Fx)^2 + (sum of all Fy)^2)^0.5

Please let me know if you still have any doubts.
 
According to your figure, my x-axis would be your A and my y-axis would be your negative-E direction.

And find the angle that each force each force is inclined with A in the counter clockwise direction from A. For example B makes 60' with A, C makes 135', D makes 210' and E makes 270' with A. So that is the 't' for each force in my equation.
 
LJX_Sham said:
According to your figure, my x-axis would be your A and my y-axis would be your negative-E direction.

And find the angle that each force each force is inclined with A in the counter clockwise direction from A. For example B makes 60' with A, C makes 135', D makes 210' and E makes 270' with A. So that is the 't' for each force in my equation.

what i mean is how to know if when should i use the sin and cos function in computing the components of x and y :(( - sorry :(( i didnt get it :((
 
  • #10
Whenever the forces are in an arbitrary direction, you will have to use sin and cos to resolve it into two convenient directions... preferably x and y..
 
  • #11
LJX_Sham said:
Whenever the forces are in an arbitrary direction, you will have to use sin and cos to resolve it into two convenient directions... preferably x and y..

:( yeah we will use sin and cos. but my prob. is how to determine if when to use it in my 1st post kindly take a look in x solving for the x component they use sin and cos. my prob is that . how can i determine the function to be use
 
  • #12
I had gone for lunch..

That is simply because your prof has used a different angle from what we have discussed.

The angle B makes is 30 with vertical. Mind it. That is 60 with horizontal..
so it should be Fcos60 which is also same as F sin30 = F cos(90-30)

As a thumb rule, to avoid confusion, find the angle that each force makes with the horizontal, and then use cos for x and sin for y.
 
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