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.