1. forces

A force of 40N acts at 35o to the horizontal together with a force of 25N acting at 110o to the horizontal. By considering force components OR by the triangle of forces determine the magnitude and direction of the resultant force.

2. Originally Posted by abz
A force of 40N acts at 35o to the horizontal together with a force of 25N acting at 110o to the horizontal. By considering force components OR by the triangle of forces determine the magnitude and direction of the resultant force.
1. If you use the parallelogram of forces you must calculate the length of a diagonal of the parallelogram. That is a side of a triangle of which you know 2 sides and the included angle = 180° - (110° - 35°) = 105°
Then use cosine rule

2. Consider the forces as vectors:

$\overrightarrow{F_1} = 40\ N(\cos(35^\circ)~, ~\sin(35^\circ))$

$\overrightarrow{F_2} = 25\ N(\cos(110^\circ)~, ~\sin(110^\circ))$

Now use vector addition to calculate the resulting force.

Use the components of the result vector to calculate the angle to the horizontal.

3. thanks mate but i dont understand what i need 2 do after that.

4. Originally Posted by abz
thanks mate but i dont understand what i need 2 do after that.
Which method do you want to perform? #1 or #2?

5. method 2 seems to be the easier of the two lets go for that.

6. Originally Posted by abz
method 2 seems to be the easier of the two lets go for that.
OK. As you may know the resulting vector is the sum of the two single forces:

$\overrightarrow{F_R} = \overrightarrow{F_1} +\overrightarrow{F_2}$

Plug in the terms of $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ of my previous post to calculate the components of $\overrightarrow{F_R}$

$\overrightarrow{F_R} = (40\ N \cdot \cos(35^\circ) + 25\ N \cdot \cos(110^\circ)~,~ 40\ N \cdot \sin(35^\circ) + 25\ N \cdot \sin(110^\circ)$

Let h and v denote the horizontal or vertical component of $\overrightarrow{F_R}$ then the magnitude of the resulting force is:

$|\overrightarrow{F_R}| = \sqrt{h^2 + v^2} \approx 52.37\ N$

The angle between the horizontal and $\overrightarrow{F_R}$ can be calculated by:

$\alpha = \arctan\left(\frac vh \right) \approx 62.458^\circ$

7. i have tried doing the above but i am left confused.
i thought the question was asking me to calculate the magnitude and the direction of the resultant force?

8. Originally Posted by abz
i have tried doing the above but i am left confused.
i thought the question was asking me to calculate the magnitude and the direction of the resultant force?
That's what earboth has given you: the calculations and result of adding two vectors by the component method.

Are you, perhaps, using a different method? Please let us see either your work on the problem or describe an example, then we might be able to help you better.

-Dan

9. i dont no what im doing any more. i understand that the answer is there but i dont understand the working to get there. no point in me copyin the answer down.

10. Abz, each force F applied at an angle has two components:
1. The horizontal component (denoted $F_{x}$)
2. The vertical component (denoted $F_{y}$)

Now, what we are trying to do is to find the sum of the horizontal components ( $F_{x} = F_{x1} + F_{x2}$) and vertical components ( $F_{y} = F_{y1} + F_{y2}$) of both of the given forces. Once we have these, we apply pythagorean theorem ( $(F_{x})^2 +(F_{y})^2 = F^2$) to find the resultant force. This is what earboth has been doing. Do you have a problem finding $F_{x}$ or $F_{y}$ ? Do you understand what we're trying to do?

11. ohhhhhhhhhhhhhhhhhh i get it. thank you it was just the wording of it. thanks.

12. Originally Posted by abz
i have tried doing the above but i am left confused.
i thought the question was asking me to calculate the magnitude and the direction of the resultant force?
I've attached the sketch I used to do your problem. Maybe this will help you to understand my calculations.