Question

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

Answer 1

`\theta=5.71^(o)`

Step-by-step explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

`\sum F_(y)=0`

`-W_(y)+N=0`

`N=W_(y)`

so:

`N=mgcos(\theta)`

now we can go ahead and do a sum of forces in the x-direction:

`\sum F_(x)=0`

the sum of forces in x is 0 because it's moving at a constant speed.

`-f+W_(x)=0`

`-\mu_(k)N+mg sen(\theta)=0`

`-\mu_(k)mg cos(\theta)+mg sen(\theta)=0`

so now we solve for theta. We can start by factoring mg so we get:

`mg(-\mu_(k) cos(\theta)+sen(\theta))=0`

we can divide both sides into mg so we get:

`-\mu_(k) cos(\theta)+sen(\theta)=0`

this tells us that the problem is independent of the mass of the object.

`\mu_(k) cos(\theta)=sen(\theta)`

we now divide both sides of the equation into
`cos(\theta)` so we get:

`\mu_(k)=(sen(\theta))/(cos(\theta))`

`\mu_(k)=tan(\theta)`

so we now take the inverse function of tan to get:

`\theta=tan^(-1)(\mu_(k))`

so now we can find our angle:

`\theta=tan^(-1)(0.10)`

so

`\theta=5.71^(o)`