Finding resultant force with no vertical component

Question

I have the question "Resolve the forces acting on the following objects into their vertical and horizontal components and thus find the vertical and horizontal components of their resultant force."

Here the vertical component is not given. Only the horizontal is given.

Here is my attempt. I have created a triangle and have used trigonometry to try and find the horizontal and vertical components:

Here I got 3.46 N for the vertical component and 2.0N for the horizontal component. However the solutions say that the resultant force should be 8.0 N.

I am not sure how this is achieved.

Answer 1

Let three forces F1, F2 and F3.

F1 = 4 cos 60° = 2 N

F2 = 4 N

F3 = 4 cos 60° = 2 N

Resultant x component = F1 + F2 + F3.

Hope its help.

Answer 2

Start by defining a suitable coordinate system (this is standard practice in any physics type problem), the most convenient one is with the x and y axes passing through the mass (that is the origin is in the center of the gray circle with the positive y axis pointing up and positive x axis pointing right as usual). Label the three forces $F_1$ the upper force, $F_2$ the middle force, and $F_3$ the lower force. The resultant of a set of forces is the sum of the individual forces $$\text{Resultant Force} = F_1 + F_2 + F_3$$

However these objects are not just numbers but vectors and hence had x and y components, so in order to add them we resort to breaking them up into components (known as vector resolution). You have correctly resolved $F_1$ into it's components so we will also resolve the other two. $F_2$ makes an angle of zero with the x axis and because of this only has a horizontal component of $8$N and a vertical component of $0$N. $F_3$ is the same as $F_1$ except that it points down, it has a positive x component and a negative y component (being in the IV quadrant). You don't need to do any work for this one and can use the previous results for the first force. So the components of $F_3$ are a y component of $-3.46$N and an x component of $2$N. So to add the forces we add their respective components.

$$\text{Resultant y component} = F_{1y}+F_{2y}+ F_{3y} = 3.46\text{N} + 0\text{N} - 3.46\text{N} = 0 \text{N} $$

$$\text{Resultant x component} = F_{1x}+F_{2x}+ F_{3x} = 2\text{N} + 4\text{N} + 2\text{N} = 8 \text{N}$$

The resulting vector only has an x component and so this is the final answer