Answer:
The angle is [tex]\theta =21.8 ^o[/tex]
Explanation:
From the question we are told that
The mass of the crate is [tex]m_c = 20 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.40[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.30[/tex]
Generally for the the crate not to slip , the static frictional must be equal to the force driving the truck
i.e
[tex]F_f = F[/tex]
Now since we are considering a slope that static frictional force is mathematically represented as
[tex]F_f = mg * cos(\theta) * \mu_s[/tex]
While the force driving the truck is mathematically represented as
[tex]F = mg * sin (\theta )[/tex]
Here mg is the weight of the crate so
So
[tex]mg * cos (\theta ) \mu_s = mg * sin (\theta )[/tex]
=> [tex]\frac{sin (\theta )}{cos (\theta)} = \mu_s[/tex]
=> [tex]\theta = tan ^{-1} [\mu_s ][/tex]
=> [tex]\theta = tan ^{-1} [0.40 ][/tex]
=> [tex]\theta =21.8 ^o[/tex]
A diagram show an illustration is on the first uploaded image
