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)
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)
First, let's draw a diagram of the forces acting on the crate.
```
|\
| \
| \
|___\
|__|\
|| || Crate
|| ||
|| ||
-----------------------------------
Incline (35.0°)
```
The weight of the crate acts vertically downward with a force of mg, where m is the mass of the crate (2.3 kg) and g is the acceleration due to gravity (9.8 m/s²).
The force of static friction acts parallel to the incline and opposes the tendency of the crate to slide down. Its magnitude is given by fs = μs * N, where μs is the coefficient of static friction and N is the normal force.
The normal force N is the component of the weight perpendicular to the incline, and it is given by N = mg * cos(θ), where θ is the incline angle (35.0°).
Therefore, the magnitude of the force of static friction is fs = μs * mg * cos(θ).
To prevent the crate from sliding down the incline, the force F applied perpendicular to the incline must be equal to or greater than the force of static friction fs.
Thus, we can write the equation:
F ≥ fs
Substituting the known values:
F ≥ μs * mg * cos(θ)
Now, we can plug in the given values to find the minimum force required to prevent the crate from sliding down the incline.
μs = 0.241 (coefficient of static friction)
m = 2.3 kg (mass of the crate)
g = 9.8 m/s² (acceleration due to gravity)
θ = 35.0° (incline angle)
F ≥ 0.241 * 2.3 kg * 9.8 m/s² * cos(35.0°)
To find the value of cos(35.0°), you can use a scientific calculator or an online calculator.
Once you have the value of cos(35.0°), you can multiply it by the other known values and calculate the minimum force required to prevent the crate from sliding down the incline.