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What is  F⃗ 3
? Give your answer in component form.

To find the vector \(\vec{F}_3\) in component form, we need to first analyze the given forces in the diagram. 

- Force \(\vec{F}_1 = 5.0 \, \text{N}\) is directed downward along the negative \(y\)-axis.
- Force \(\vec{F}_2 = 6.0 \, \text{N}\) is directed at an angle of 20° from the positive \(x\)-axis.
- Force \(\vec{F}_3\) is pointing upward along the positive \(y\)-axis (which we need to find in component form).
- Force \(\vec{F}_4\) points to the left along the negative \(x\)-axis.

### Step 1: Decompose the known forces into components
We can decompose \(\vec{F}_2\) into its \(x\) and \(y\) components using trigonometry:
- \(F_{2x} = 6.0 \, \text{N} \cdot \cos(20^\circ)\)
- \(F_{2y} = 6.0 \, \text{N} \cdot \sin(20^\circ)\)

We already know that:
- \(\vec{F}_1\) has components \((0, -5.0 \, \text{N})\).
- \(\vec{F}_4\) has components \((-F_4, 0)\), with \(F_4\) being the magnitude of the force.

### Step 2: Apply equilibrium conditions
Since the system appears to be in equilibrium (the forces sum to zero), we can use Newton's first law to set up the equations:
- In the \(x\)-direction: \(\sum F_x = 0\)
- In the \(y\)-direction: \(\sum F_y = 0\)

Let's start by solving for \(\vec{F}_3\). I'll compute the components of \(\vec{F}_2\) and then balance the forces to find \(\vec{F}_3\).The components of \(\vec{F}_2\) are:
- \(F_{2x} = 5.64 \, \text{N}\) (in the positive \(x\)-direction)
- \(F_{2y} = 2.05 \, \text{N}\) (in the positive \(y\)-direction)

The component of \(\vec{F}_1\) in the \(y\)-direction is:
- \(F_{1y} = -5.0 \, \text{N}\) (in the negative \(y\)-direction)

### Step 3: Balance forces in the \(y\)-direction
For equilibrium in the \(y\)-direction, the sum of all vertical forces must equal zero:
\[
F_{3y} + F_{1y} + F_{2y} = 0
\]
\[
F_{3y} = - (F_{1y} + F_{2y})
\]

Now let's calculate \(F_{3y}\), which is the magnitude of \(\vec{F}_3\).The magnitude of \(\vec{F}_3\) in the positive \(y\)-direction is approximately \(2.95 \, \text{N}\).

Thus, in component form, \(\vec{F}_3 = (0, 2.95 \, \text{N})\), since it only acts along the \(y\)-axis.

Would you like more details, or have any other questions?

Here are 5 related questions you could explore:
1. What is the magnitude of force \(\vec{F}_4\)?
2. How do you determine the net force in the \(x\)-direction?
3. How would the equilibrium change if force \(\vec{F}_2\) had a different angle?
4. What would happen if an additional force was introduced along the positive \(y\)-axis?
5. Can you calculate the resultant force from all the forces shown?

**Tip:** In equilibrium problems, always separate forces into components before applying equilibrium conditions.

What is F⃗ 3? Give your answer in component form.

Learn how to find the component form of vector F3 in an equilibrium problem with forces acting along different axes. Using trigonometry and Newton's First Law, we decompose the forces and solve for F3. Suitable for Grades 10-12.

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