I know that the answer to this problem is (0.2237232293) and to begin it by making a force body diagram, but I need help setting up the equation(s). Thanks!
A box rest on an incline making a 30 angle with the horizontal. It is found that a parallel force to the incline of at least 245 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.
Based on the information given, 245 N. is too low. Please check all of the given INFO for accuracy.
To find the coefficient of static friction between the box and the incline, we can start by analyzing the forces acting on the box.
First, let's draw a force diagram for the box:
^ Normal Force (N)
|
-----|----- Weight of the Box (800 N)
|
F | (30°)
═════|═════ Incline
|
|
We have the weight of the box acting vertically downward with a magnitude of 800 N. The incline makes a 30° angle with the horizontal. We need to find the coefficient of static friction, which is denoted by μ (mu).
The normal force (N) acts perpendicular to the incline. It can be calculated as N = m × g, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s²). Since we know the weight (W) of the box, we can use W = m × g to solve for m. Rearranging the equation gives m = W / g.
Substituting the given values, m = 800 N / 9.8 m/s² ≈ 81.63 kg.
To prevent the box from sliding down the incline, a parallel force (F) of at least 245 N is required. This force can be seen as the static friction force acting up the incline.
The static friction force (f_s) can be calculated as f_s = μ × N, where μ is the coefficient of static friction. Rearranging the equation gives μ = f_s / N.
We know that f_s must be at least 245 N to prevent sliding, so we can substitute this value into the equation. Also, the previously calculated value of N can be substituted:
μ = 245 N / (81.63 kg × 9.8 m/s²) ≈ 0.301
Therefore, the coefficient of static friction between the box and the incline is approximately 0.301.