Step-by-Step Solution

We are given:

• Mass of the child and sled system: $m = 31.5 \, \text{kg}$
• Magnitude of the pushing force: $F_{\text{push}} = 42 \, \text{N}$
• Angle of the applied force: $\theta = 21.10^\circ$
• The system is moving at a constant velocity.

Since the system is moving at a constant velocity, the net force in both horizontal and vertical directions must be zero, meaning all forces balance each other.

Forces Acting on the System

1. Gravitational Force ($F_g$)
   The weight of the sled-child system is given by: $F_g = m \cdot g = 31.5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 308.415 \, \text{N}$ This acts vertically downward.

2. Normal Force ($F_N$)
   This is the upward force exerted by the ground on the sled. It balances out the vertical forces.

3. Pushing Force ($F_{\text{push}}$)
   The applied force has two components:

   • Horizontal component: $F_{\text{push},x} = F_{\text{push}} \cdot \cos(\theta)$
   • Vertical component: $F_{\text{push},y} = F_{\text{push}} \cdot \sin(\theta)$

4. Frictional Force ($F_f$)
   The frictional force opposes the motion and acts horizontally. Since the sled moves at constant velocity, the horizontal component of the applied force must equal the frictional force.

Free-Body Diagram

We can summarize the forces as follows:

• Vertical direction:

  • $F_N + F_{\text{push},y} = F_g$
  • From this, we can solve for the normal force $F_N$.

• Horizontal direction:

  • $F_{\text{push},x} = F_f$
  • This will allow us to solve for the frictional force.

Horizontal Force Balance

The horizontal component of the applied force is: $F_{\text{push},x} = F_{\text{push}} \cdot \cos(\theta)$ Substituting the given values: $F_{\text{push},x} = 42 \, \text{N} \cdot \cos(21.10^\circ) = 42 \, \text{N} \cdot 0.933 = 39.186 \, \text{N}$ Since the system is moving at a constant velocity, the frictional force must equal the horizontal component of the pushing force: $F_f = 39.186 \, \text{N}$

Vertical Force Balance

In the vertical direction: $F_N + F_{\text{push},y} = F_g$ where the vertical component of the applied force is: $F_{\text{push},y} = F_{\text{push}} \cdot \sin(\theta)$ Substituting the values: $F_{\text{push},y} = 42 \, \text{N} \cdot \sin(21.10^\circ) = 42 \, \text{N} \cdot 0.360 = 15.120 \, \text{N}$ Now, solving for the normal force: $F_N = F_g - F_{\text{push},y} = 308.415 \, \text{N} - 15.120 \, \text{N} = 293.295 \, \text{N}$

Final Answer

The frictional force is approximately: $F_f = 39.2 \, \text{N}$

Would you like a detailed drawing of the free-body diagram or any further explanation?

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Here are 5 related questions you might find interesting:

1. How does the angle of the applied force affect the normal force?
2. What happens to the frictional force if the surface is inclined?
3. How would the frictional force change if the mass of the child-sled system increased?
4. How is the coefficient of friction related to the normal and frictional forces?
5. What role does kinetic friction play in a system moving at constant velocity?

Tip: Always break forces into components when dealing with angles to simplify your calculations!