# Calculating two diagonal forces

When I tried to calculate the force exerted on the rope 'a' I get that it is equal to 5N when the block has a force of 10N. Do I instead need to divide by 30 since the sum of the two angles is 30 degrees? Should I be using the cosine of the angles so that it is (cos(10) / cos(20)) * 10? I just need to calculate the force exerted on rope 'a'. Thanks.

• The tag kinematics is not applicable here. This is just Newton's 3rd law. Oct 2 at 16:22

What you need to consider is the equilibrium of the body.

On the vertical axis Y, there is the F=10 N force, and the vertical components from the forces of the ropes, ie:

• $$F_{A,y}$$: the vertical component of the force from rope A
• $$F_{B,y}$$: the vertical component of the force from rope B

Additionally on the X-axis (horizontal), you only have two forces

• $$F_{A,x}$$: the horizontal component of the force from rope A
• $$F_{B,x}$$: the horizontal component of the force from rope B

The forces are presented in the following graph

In both cases the sum of the forces in each axis should be zero. i.e.:

• $$\sum F_x =0 \Rightarrow -F_{A,x}+ F_{B,x}=0$$
• $$\sum F_y =0 \Rightarrow F - F_{A,y}- F_{B,y}=0$$

Also, the relationship between the components of the forces and the force on the rope will follow the following relationship .

$$F_{A,x} = F_{A}\cdot \sin(\theta_a) \qquad F_{A,y} = F_{A}\cdot \cos(\theta_a)$$

I only wrote for rope A, because I think this looks like homework. Although you have showed what you've tried, its far away from being correct, so I tried to put you in the right track. Hopefully you will be able to find the solution. I'd be happy to assist more if you update your solution.

There is a piece of important information missing from the problem statement - whether the position shown represents the final equilibrate state with the assumptions that the angles and the length of the ropes remain constant, thus no change in the shape of either triangle (see sketch below).

If this is the case, then you can solve the question simply by engineering mechanics, $$\sum F_x = 0$$ (since the apex will not move), and $$\sum F_y = 0$$), that is:

$$T_A cos\theta_A = T_B cos\theta_B$$, and $$T_A sin\theta_A + T_B sin\theta_B = T$$

However, if the assumption are not true, then you need to take into account of the deformation of the ropes (geometry changes) as shown on the sketch. The equilibrium equations remain the same but with the terms having a superscription "$$'$$" - $$T_A cos\theta_A' = T_B cos\theta_B'$$, and $$T_A sin\theta_A' + T_B sin\theta_B' = T$$, and with an additional assumption, that for change of angle is small, the distances of points $$A$$ - $$C$$, and $$C$$ - $$B$$ are approximately constant before and after the stretches due to the applied load.

• I'm curious what software was used to create this sketch? Sep 7, 2022 at 2:17
• #MeToo 87654321 Jan 5 at 1:57
• Using any graphic program such as cad. This one was using Openoffice's drawing tool.
– r13
Jan 5 at 2:46
• Some alternatives: Paint.net (bitmaps with layers), Geogebra (used in schools), Tikz (hardly anything you can‘t visualize with it). May 5 at 4:11

Here's a new answer to an old question.

## Oversimplification

Assuming $$10° \approx 20° \approx 0°$$, the forces $$F_A$$ and $$F_B$$ in both ropes are equal as both ropes are in parallel, hence $$5$$ N each. However, this oversimplification will not hold: it's just a limiting case.

## Calculations

The diagram shows the ropes in its lower part (black) and the force-vectors in its upper part (red). We can derive two equations from it in two variables, which we can solve.

For the resultant in y-direction we have:

$$F_A \cdot \cos(10°) + F_B \cdot \cos(20°) = 10 N$$

While in x-direction (vector $$\vec F_c$$) this equality must hold:

$$F_A \cdot \sin(10°) = F_B \cdot \sin(20°)$$

or

$$F_B = F_A \cdot \frac{\sin(10°)}{\sin(20°)}$$

which yields by substitution

$$F_A \cdot \cos(10°) + F_A \cdot \frac{\sin(10°)}{\sin(20°)} \cdot \cos(20°) = 10 N$$

or

$$F_A = 10 N \cdot \frac{1}{\cos(10°) + \frac{\sin(10°)}{\tan(20°)}} = 6.8 N$$

As can be seen easily, e.g. by Taylor approximation, the fraction approaches $$\frac{1}{2}$$ when both angles approach $$0°$$, as we noted in the oversimplification ($$\cos(x) \approx1$$, $$\sin(x) \approx x$$, $$\tan(x) \approx x$$, [$$x$$] = 1 rad, $$x \rightarrow 0$$).

(done by tikz)