A 3.66 kg block is placed on an inclined surface that makes an angle of 18.5° relative to horizontal. The coefficients of static and kinetic friction between the block and surface are 0.534 and 0.239, respectively. The block remains at rest when released. What is the magnitude of the static frictional force acting on the block?
F(fr) = k(st) •N = k(st)•m•g•cosα
To find the magnitude of the static frictional force acting on the block, we can use the following equation:
Fs ≤ μs * N
Where:
Fs = static frictional force
μs = coefficient of static friction
N = normal force
First, let's calculate the normal force acting on the block. It can be determined by decomposing the weight of the block along the inclined surface.
N = mg * cos(θ)
Where:
m = mass of the block
g = acceleration due to gravity
θ = angle of the inclined surface
Given information:
m = 3.66 kg
g ≈ 9.8 m/s²
θ = 18.5°
Calculating the normal force:
N = (3.66 kg) * (9.8 m/s²) * cos(18.5°)
Now, let's calculate the static frictional force:
Fs ≤ (0.534) * N
Substituting the value of N:
Fs ≤ (0.534) * [(3.66 kg) * (9.8 m/s²) * cos(18.5°)]
You can input the above equation into a calculator or use a trigonometric function to evaluate cos(18.5°). The result will give you the maximum value of static friction, Fs. This maximum value represents the magnitude of the static frictional force acting on the block when it is at rest.