if a 900-kg box slides down in a slide inclined at 25degrees to the horizontal and the coefficient of friction between the box and the slide is .0500, what is the friction force?
To find the friction force acting on the box as it slides down the inclined slide, we can use the following steps:
1. Identify the forces involved:
- Weight (W): The force due to the mass of the box, acting vertically downward.
- Normal force (N): The force perpendicular to the surface of the slide, counteracting the weight.
- Friction force (f): The resistive force opposing the motion of the box along the slide.
- Force parallel to the slide (F_parallel): The component of weight acting down the incline in the direction of motion.
2. Calculate the weight (W) of the box:
Weight (W) = mass (m) * acceleration due to gravity (g).
Given mass = 900 kg and g = 9.8 m/s², we can find:
W = 900 kg * 9.8 m/s² = 8820 N.
3. Calculate the force parallel to the slide (F_parallel):
F_parallel = W * sin(angle of inclination).
The angle of inclination is given as 25 degrees, so we have:
F_parallel = 8820 N * sin(25°) = 3728.66 N.
4. Calculate the normal force (N) on the box:
N = W * cos(angle of inclination).
Using the same angle of inclination of 25 degrees, we have:
N = 8820 N * cos(25°) = 7985.94 N.
5. Calculate the maximum static friction force (f_max):
f_max = coefficient of friction (µ) * N.
Given the coefficient of friction (µ) = 0.0500, we can calculate:
f_max = 0.0500 * 7985.94 N = 399.30 N.
6. Determine if the friction force (f) is equal to or less than the maximum static friction force (f_max):
Since the box is sliding down the slope, the friction force will be less than the maximum static friction force.
7. So, the friction force (f) will be equal to the force parallel to the slide (F_parallel):
f = F_parallel = 3728.66 N.
Therefore, the friction force acting on the box as it slides down the inclined slide is 3728.66 N.