Let's analyze the provided problem step-by-step based on the given diagram and forces:

Part (a): Find the direction and magnitude of the total force $F_{tot}$:

We are given:

• $F_1 = 29.6 \, \text{N}$
• $F_2 = 18.2 \, \text{N}$

These forces act at right angles to each other, with $F_1$ acting along the horizontal (x-axis) and $F_2$ along the vertical (y-axis).

1. Magnitude of the total force $F_{tot}$: The resultant force can be found using the Pythagorean theorem since the forces are perpendicular: $F_{tot} = \sqrt{F_1^2 + F_2^2}$ $F_{tot} = \sqrt{(29.6)^2 + (18.2)^2}$ $F_{tot} = \sqrt{876.16 + 331.24} = \sqrt{1207.4} \approx 34.75 \, \text{N}$

2. Direction of $F_{tot}$: The direction of $F_{tot}$ is given by the angle $\theta$ with respect to the horizontal direction of $F_1$. This angle can be calculated using the tangent function: $\theta = \tan^{-1}\left( \frac{F_2}{F_1} \right)$ $\theta = \tan^{-1}\left( \frac{18.2}{29.6} \right)$ $\theta \approx \tan^{-1}(0.614) \approx 31.68^\circ$

Thus, the magnitude of the total force $F_{tot}$ is 34.75 N, and its direction is approximately 31.68° counterclockwise from the direction of $F_1$.

Part (b): Initial acceleration if stationary (with static friction):

The friction force needs to be overcome for the skater to start moving. We use the formula: $F_{\text{friction}} = \mu_s \cdot m \cdot g$ Where:

• $\mu_s = 0.04$ (coefficient of static friction for steel on ice)
• $m = 65.0 \, \text{kg}$ (mass of the skater)
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)

Calculating the friction force: $F_{\text{friction}} = 0.04 \cdot 65.0 \cdot 9.81 \approx 25.5 \, \text{N}$

Since the total applied force $F_{tot} = 34.75 \, \text{N}$ is greater than the static friction $F_{\text{friction}} = 25.5 \, \text{N}$, the skater will begin to move.

The initial acceleration $a$ can be found using Newton's second law: $a = \frac{F_{\text{net}}}{m} = \frac{F_{tot} - F_{\text{friction}}}{m}$ $a = \frac{34.75 \, \text{N} - 25.5 \, \text{N}}{65.0 \, \text{kg}} \approx \frac{9.25}{65} \approx 0.14 \, \text{m/s}^2$

Part (c): Acceleration assuming motion (with kinetic friction):

For motion, the friction is kinetic, not static. If we assume the same coefficient of kinetic friction as static (which is likely in this case), the kinetic friction force is: $F_{\text{friction, kinetic}} = \mu_k \cdot m \cdot g = 0.04 \cdot 65.0 \cdot 9.81 \approx 25.5 \, \text{N}$

The acceleration will still be: $a = \frac{F_{tot} - F_{\text{friction, kinetic}}}{m}$ $a = \frac{34.75 \, \text{N} - 25.5 \, \text{N}}{65.0 \, \text{kg}} \approx 0.14 \, \text{m/s}^2$

Thus, the acceleration remains approximately 0.14 m/s² even when the skater is in motion.

Would you like more details on any of these calculations, or have any other questions?

Here are five follow-up questions you might find useful:

1. What happens to the acceleration if the coefficient of friction is reduced further?
2. How does the skater's mass affect the total force required to overcome friction?
3. What role does the angle of force application play in determining acceleration?
4. How would the results change if the skater were on a different surface with a higher coefficient of friction?
5. Can we calculate the work done by the skaters applying forces $F_1$ and $F_2$ over a distance?

Tip: Always check the direction of the resultant force when combining vectors to ensure you’re applying trigonometric functions correctly!