Sitting on a horizontal track is a mass of 10kg. One end of the track is raised untill the mass just begins to slip. If the coeffieient of static friction is .4 what is the minimum angle at which the mass begins to slip?
The component of force down the slope:
mgSinTheta
THe component of force (friction) up the slope
mg*mu*CosTheta
Set these equal, and solve for Theta.
Check my thinking.
mgSinTheta = mg*mu*CosTheta
(10)(-9.8)sin theta = (10)(-9.8) * (.4) * cos theta
then I got -98 sin theta = -39.2 cos theta
I don't understand how to solve for theta once I got this far.
Sitting on a horizontal track is a mass of 10kg. One end of the track is raised untill the mass just begins to slip. If the coeffieient of static friction is .4 what is the minimum angle at which the mass begins to slip?
mgSinTheta = mg*mu*CosTheta
(10)(-9.8)sin theta = (10)(-9.8) * (.4) * cos theta
then I got -98 sin theta = -39.2 cos theta
I don't understand how to solve for theta once I got this far.I also wasn't sure what "mu" in the above equation ment.
mu is the coefficient of friction
To solve for theta, you can use trigonometric identities to manipulate the equation and isolate theta.
Starting with -98 sin theta = -39.2 cos theta, divide both sides of the equation by -39.2 to simplify:
(-98 sin theta) / -39.2 = cos theta
Now, recall the trigonometric identity:
sin theta / cos theta = tan theta
So, we can rewrite the equation as:
(98 / 39.2) tan theta = 1
To solve for theta, take the inverse tangent (also known as arctan) of both sides:
tan^(-1)((98 / 39.2) tan theta) = tan^(-1)(1)
This results in:
theta = tan^(-1)(1)
Using a calculator, you'll find that tan^(-1)(1) is approximately 45 degrees.
Therefore, the minimum angle at which the mass begins to slip is 45 degrees.