Answer:
[tex]\theta=5.71^{o}[/tex]
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:
[tex]\sum F_{y}=0[/tex]
[tex]-W_{y}+N=0[/tex]
[tex]N=W_{y}[/tex]
so:
[tex]N=mgcos(\theta)[/tex]
now we can go ahead and do a sum of forces in the x-direction:
[tex]\sum F_{x}=0[/tex]
the sum of forces in x is 0 because it's moving at a constant speed.
[tex]-f+W_{x}=0[/tex]
[tex]-\mu_{k}N+mg sen(\theta)=0[/tex]
[tex]-\mu_{k}mg cos(\theta)+mg sen(\theta)=0[/tex]
so now we solve for theta. We can start by factoring mg so we get:
[tex]mg(-\mu_{k} cos(\theta)+sen(\theta))=0[/tex]
we can divide both sides into mg so we get:
[tex]-\mu_{k} cos(\theta)+sen(\theta)=0[/tex]
this tells us that the problem is independent of the mass of the object.
[tex]\mu_{k} cos(\theta)=sen(\theta)[/tex]
we now divide both sides of the equation into [tex]cos(\theta)[/tex] so we get:
[tex]\mu_{k}=\frac{sen(\theta)}{cos(\theta)}[/tex]
[tex]\mu_{k}=tan(\theta)[/tex]
so we now take the inverse function of tan to get:
[tex]\theta=tan^{-1}(\mu_{k})[/tex]
so now we can find our angle:
[tex]\theta=tan^{-1}(0.10)[/tex]
so
[tex]\theta=5.71^{o}[/tex]
