The coefficient of static friction between the 2.3 kg crate and the 35.0° incline of Figure P4.47 is 0.241. What minimum force, F, must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
ΣF(x) = 0: m•g•sinα=F(fr)
ΣF(y) = 0: m•g•cosα +F = N
F(fr)=μ•N= μ• (m•g•cosα +F)
m•g•sinα= μ• (m•g•cosα +F)
F=(m•g/μ) •(sinα -μ• cosα)
To determine the minimum force required to prevent the crate from sliding down the incline, we need to analyze the forces acting on the crate.
First, let's break down the forces into their components. We have the weight of the crate acting vertically downward, which can be split into two components:
1. The component perpendicular to the incline (mg cosθ), where m is the mass of the crate (2.3 kg) and g is the acceleration due to gravity (9.8 m/s²).
2. The component parallel to the incline (mg sinθ).
In this case, the force perpendicular to the incline (F) counteracts the perpendicular component of the weight (mg cosθ) to prevent the crate from sliding downwards.
The force parallel to the incline (mg sinθ) can also be referred to as the force due to gravity acting along the incline. To prevent the sliding, the frictional force opposing this component must be equal to or greater than the force due to gravity along the incline.
The frictional force is given by the equation:
Frictional force (Ff) = coefficient of static friction (μ) * perpendicular force (mg cosθ)
Substituting the given values:
Ff = 0.241 * (2.3 kg * 9.8 m/s² * cos(35°))
Now, to find the minimum force that must be applied perpendicular to the incline (F), we must set F equal to Ff:
F = Ff = 0.241 * (2.3 kg * 9.8 m/s² * cos(35°))
After calculating this expression, you will find the minimum force, F, required to prevent the crate from sliding down the incline.