Question

1. procedures

a) coefficient of static friction
consider an object on a rough incline plane. as the incline in raised, the force acting down the
incline increases. more and more static friction is generated resisting the tendency for the block
to slide. at some angle the friction force is maximized and can no longer support against the
force of gravity. it is at that angle that the static friction force just balance the force of gravity.
any larger of an angle, the mass will start to slide.
(1) \\( \sum f _ { x } = m g \sin \theta - \mu _ { s } n = 0 \\)
(2) \\( \sum f _ { y } = n - m g \cos \theta = 0 \\)
giving \\( n = m g \cos \theta \\)
plug n above into equation 1 above and solve for the \\( \mu _ { s } \\) as a
function of the incline angle at which the gravitational force
along the incline equals the maximum \\( f _ { \text { static } } \\). show your
work below.

Explanation:

Step1: Substitute \(n\) into equation (1)

We know \(n = mg\cos\theta\) from equation (2). Substitute this into equation (1) \(\sum F_x=mg\sin\theta-\mu_sn = 0\). So we get \(mg\sin\theta-\mu_s(mg\cos\theta)=0\).

Step2: Solve for \(\mu_s\)

First, factor out \(mg\) from the left - hand side of the equation \(mg\sin\theta-\mu_s(mg\cos\theta)=0\). We have \(mg(\sin\theta-\mu_s\cos\theta)=0\). Since \(m
eq0\) and \(g
eq0\) (mass \(m\) of the object and acceleration due to gravity \(g\) are non - zero in a physical context), we can divide both sides of the equation by \(mg\). Then \(\sin\theta-\mu_s\cos\theta = 0\). Rearranging for \(\mu_s\), we get \(\mu_s=\frac{\sin\theta}{\cos\theta}\).

Answer:

\(\mu_s = \tan\theta\)