A 21 kg object is placed on an incline with friction. The incline is raised slowly from zero degrees and at 60.71 degrees, the object begins to slide down the incline. What is the coefficient of static friction? Be as accurate as possible with your answer and be sure to enter at least two decimal places.
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To solve this problem, we need to use the concept of equilibrium and the equation for static friction.
First, let's draw a free body diagram for the object on the incline:
/|
/ |
/θ |
W /____|
Here, W represents the weight of the object, which can be calculated as the mass (m) of the object multiplied by the acceleration due to gravity (g). In this case, the mass is given as 21 kg, and the acceleration due to gravity is approximately 9.8 m/s².
So, W = 21 kg × 9.8 m/s² = 205.8 N.
Next, let's break the weight vector into two components: one parallel to the incline (W₁) and one perpendicular to the incline (W₂):
/|
/ |
/θ |
W₂ /____|
|
|
| W₁
W₁ is calculated as W × sinθ and W₂ is calculated as W × cosθ, where θ is the angle of incline (60.71 degrees in this case).
W₁ = 205.8 N × sin(60.71°) = 178.856 N
W₂ = 205.8 N × cos(60.71°) = 102.9 N
Now, let's analyze the forces acting on the object. We have the force of gravity (W) acting downward, the normal force (N) acting perpendicular to the incline and the static friction force (f) opposing the force trying to slide the object down the incline.
Since the object is on the verge of sliding, the static friction force is at its maximum value and is equal to the component of the weight along the incline (W₁). Therefore, we can write f = μsN, where μs is the coefficient of static friction.
N can be calculated as W₂, the component of the weight perpendicular to the incline. So, N = 102.9 N.
Now, substituting the known values into the equation f = μsN, we have:
178.856 N = μs × 102.9 N
Simplifying the equation, we get:
μs = 178.856 N / 102.9 N
Calculating the value, we find:
μs ≈ 1.738
Therefore, the coefficient of static friction is approximately 1.74.