A girl (22.5 kg) goes down a straight slide inclined 38.0° above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30.0 N force as shown in the figure. How large is the normal force acting on the child?
To solve this problem, we need to analyze the forces acting on the girl and apply the principles of equilibrium. Let's break it down step by step:
1. Draw a free-body diagram: This helps us visualize the forces acting on the girl. In this case, we have the weight (mg) acting vertically downward, the normal force (N) perpendicular to the incline, the friction force (f) acting opposite to the direction of motion, and the horizontal rope force (F) acting parallel to the incline.
2. Resolve the weight force: The weight has two components. The component perpendicular to the incline (mg * cos θ) gets cancelled out by the normal force, so we only need to consider the component parallel to the incline, which is mg * sin θ.
3. Write down the equations of motion: In the vertical direction, we have N - mg * cos θ = 0, since there is no vertical acceleration. In the horizontal direction, we have F - f - mg * sin θ = 0, since there is no horizontal acceleration.
4. Solve for the unknowns: We have two equations with two unknowns (N and f). Start by solving the vertical equation for N: N = mg * cos θ. Then substitute this value into the horizontal equation to find f: F - mg * sin θ = f.
5. Calculate the normal force: Now that we have found the value of N, we can substitute it back into the vertical equation: N = mg * cos θ.
N = (22.5 kg)(9.8 m/s^2)(cos 38.0°)
Make sure to convert the angle to radians: θ = 38.0° * (π/180°)
N ≈ 199.49 N
Therefore, the normal force acting on the child is approximately 199.49 N.