A 38 kg child is sitting on a slide at angle 45o from the horizontal. What is the minimum value of static friction between the child and the slide that prevents the child from sliding down?
Wc = m*g = 38kg * 9.8N/kg = 372.4 N. =
Wt. of child.
Fc = 372.4 N. @ 45o. = Force of child.
Fp = 372.4*sin45 = 263.3 N. = Force
parallel to slide.
Fv = 372.4*cos45 = 263.3 N. = Force
perpendicular to slide.
Fs = Force of static friction.
Fp-Fs = m*a.
372.4 - Fs = 38*0 = 0
Fs = 372.4 N. = Min.static friction.
Correction:
Fp-Fs = m*a
263.3 - Fs = 38*0 = 0
Fs = 263.3 N.=Force of static friction.
To find the minimum value of static friction between the child and the slide, we need to consider the forces acting on the child.
1. Identify the forces: In this case, the forces acting on the child are the gravitational force (mg) pulling downwards vertically, and the static friction force (f) opposing the motion along the slide.
2. Resolve the forces: Since the slide makes an angle of 45 degrees with the horizontal, we need to resolve them into components parallel and perpendicular to the slide.
- The weight (mg) can be broken down into its components as follows:
- The component parallel to the slide is mg * sin(45°) since sin(45°) = √2 / 2.
- The component perpendicular to the slide is mg * cos(45°) since cos(45°) = √2 / 2.
3. Equate the forces: For the child to remain in equilibrium (not sliding down or up), the static friction force should be equal to the component parallel to the slide.
- So, the equation becomes f = mg * sin(45°).
4. Calculate the minimum static friction force: Plug in the given mass (m) and acceleration due to gravity (g = 9.8 m/s^2) to find the minimum static friction value.
- f = 38 kg * 9.8 m/s^2 * (√2 / 2)
- f ≈ 270.4 N
Therefore, the minimum value of the static friction between the child and the slide that prevents the child from sliding down is approximately 270.4 Newtons.