Let's solve for the magnitude of the normal force $F_n$ in terms of the weight $F_w$ and the angle $\theta$, by eliminating $F_f$ from the equations for the sum of forces in the x and y directions.

Step 1: Equations for the sum of forces

From the previous parts, we have the following two equations:

\sum F_y = 0 = F_n \cos(\theta) + F_f \sin(\theta) - F_w \tag{1}

\sum F_x = 0 = -F_n \sin(\theta) + F_f \cos(\theta) \tag{2}

Step 2: Eliminate $F_f$

To eliminate $F_f$, we multiply Equation (1) by $\cos(\theta)$ and Equation (2) by $\sin(\theta)$:

• Multiply Equation (1) by $\cos(\theta)$:

F_n \cos^2(\theta) + F_f \sin(\theta) \cos(\theta) = F_w \cos(\theta) \tag{3}

• Multiply Equation (2) by $\sin(\theta)$:

-F_n \sin^2(\theta) + F_f \sin(\theta) \cos(\theta) = 0 \tag{4}

Step 3: Subtract the equations

Now, subtract Equation (4) from Equation (3) to cancel out the $F_f \sin(\theta) \cos(\theta)$ terms:

$F_n \cos^2(\theta) + F_n \sin^2(\theta) = F_w \cos(\theta)$

Step 4: Use the trigonometric identity

Since $\cos^2(\theta) + \sin^2(\theta) = 1$, the equation simplifies to:

$F_n = F_w \cos(\theta)$

Thus, the magnitude of the normal force is:

$F_n = F_w \cos(\theta)$

This expression gives the normal force $F_n$ in terms of the weight $F_w$ and the angle $\theta$, with no dependence on the frictional force $F_f$.

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Do you want further details or have any questions about this solution?

Here are 5 related questions to explore further:

1. How would the expression for $F_n$ change if there was vertical motion?
2. What role does the angle $\theta$ play in determining the frictional force?
3. How does the normal force affect the frictional force on the block?
4. How would the solution change if the block were on an inclined plane?
5. Can this method be applied to a three-dimensional force system?

Tip: When working with trigonometric identities in physics, it helps to simplify expressions early using identities like $\sin^2(\theta) + \cos^2(\theta) = 1$.