Question (with solution) which I don't understand (statics)

[Riazy]

Homework Statement

Calculation of the forces C and T in the figure

http://www.picpaste.com/20140127_171733-lxIN0kJQ.jpg
http://www.picpaste.com/20140127_171733-lxIN0kJQ.jpg

My problem is that I don't understand how cos and sin is given
I know the rules SOHCAHTOA

Homework Equations

Equations used are
Sum of Fx= 0
and Sum of Fy = 0

The Attempt at a Solution

As this is an example from my coursework material, it is already solved. But I want to understand it there for I will present the solution

Sum of Fx = 0

8+Tcos40 + Csin20 - 16 = 0
0,766T+0,342C=8

Sum of Fy = 0

Tsin40-Ccos20-3=0
0,643T-0,940C = 3

By combining the both equations we can solve for C and T

Answer is T = 9,09 kn and C = 3,03 kn

So basically my question is how are cos and sin given? yeah i know its a triangle
and I know of SOH CAH TOA rule.. but i don't seem to grasp it when in this Picture

Actually as I am used to see a problem I first

draw a triangle for example for the 40 degree one, sometimes i divide a resultant force into two composants

I see that T = hypotenuse
Then I look at the adjacent side, and normally there is a variable there which is present without a value that I want to be solved.. Now in this problem 8 isn't used as a value. Normally I would divide T into Tx and Ty, and draw to triangles.


So how should I look at this problem now?

how do i know its sin i will use respectively cos... if someone could explain it It would be very kind of you . how should I Think? B]

I Believe in this problem I am not supposed to divide into two composants, but i want to know how they decide "when to use cos resp sin"

 

[SteamKing]

You have to master which side of the angle is adjacent to the angle and which is opposite the angle.

For vector T, the adjacent side of the 40 degree angle is the x-axis, so the horizontal component is T cos 40.

For vector C, the adjacent side of the 20 degree angle is the y-axis, so the vertical component is C cos 20.

 

[Riazy]

Sum of Fx = 0

8+Tcos40 + Csin20 - 16 = 0
0,766T+0,342C=8

Sum of Fy = 0

Tsin40-Ccos20-3=0
0,643T-0,940C = 3


could you show me somehow?

 

[SteamKing]

Draw out the triangles on the diagram. That way, you will be able to see how the sin and cos relationships correspond to the horizontal and vertical components of the different vectors.

 

[rezeew]

and how they set up the equation as they did.

As a scientist, it is important to understand the fundamentals of trigonometry in order to solve problems in statics. In this particular problem, the use of cos and sin is determined by the angle between the forces and the horizontal and vertical axes. Let's break down the problem to better understand this.

First, we have two forces acting on the object: T and C. T is acting at an angle of 40 degrees from the horizontal axis and C is acting at an angle of 20 degrees from the vertical axis. These angles are important because they will determine which trigonometric functions (cos or sin) we will use in our equations.

Next, we need to draw a free body diagram of the object. This will help us visualize the problem and identify all the forces acting on the object. In this case, we can draw a triangle for each force, with the hypotenuse representing the magnitude of the force and the adjacent side representing the component of the force in the respective direction (horizontal or vertical).

Now, let's look at the equations used in the solution. The first equation, Sum of Fx = 0, is used to find the horizontal component of the forces. In this case, we have Tcos40 and Csin20. The reason we use cos for Tcos40 is because we are looking at the horizontal component of T, and cos represents the adjacent side over the hypotenuse in a right triangle. Similarly, we use sin for Csin20 because we are looking at the vertical component of C, and sin represents the opposite side over the hypotenuse in a right triangle.

The second equation, Sum of Fy = 0, is used to find the vertical component of the forces. In this case, we have Tsin40 and Ccos20. Again, we use sin for Tsin40 because we are looking at the vertical component of T, and sin represents the opposite side over the hypotenuse in a right triangle. Similarly, we use cos for Ccos20 because we are looking at the horizontal component of C, and cos represents the adjacent side over the hypotenuse in a right triangle.

By combining these two equations, we can solve for the unknown forces T and C. It is important to note that the use of cos and sin in these equations is determined by the angles between the forces and the horizontal and vertical axes.

In summary, when