Question

A flatbed truck is carrying a 20-kg crate up a sloping road. The coefficient of static friction between the crate and the bed is 0.40, and the coefficient of kinetic friction is 0.30. What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place

Answer 1

The angle is
`\theta =21.8 ^o`

Step-by-step explanation:

From the question we are told that

The mass of the crate is
`m_c = 20 \ kg`

The coefficient of static friction is
`\mu_s = 0.40`

The coefficient of kinetic friction is
`\mu_k = 0.30`

Generally for the the crate not to slip , the static frictional must be equal to the force driving the truck

i.e

`F_f = F`

Now since we are considering a slope that static frictional force is mathematically represented as

`F_f = mg * cos(\theta) * \mu_s`

While the force driving the truck is mathematically represented as

`F = mg * sin (\theta )`

Here mg is the weight of the crate so

So

`mg * cos (\theta ) \mu_s = mg * sin (\theta )`

=>
`(sin (\theta ))/(cos (\theta)) = \mu_s`

=>
`\theta = tan ^(-1) [\mu_s ]`

=>
`\theta = tan ^(-1) [0.40 ]`

=>
`\theta =21.8 ^o`

A diagram show an illustration is on the first uploaded image