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.
-98 sin theta = -39.2 cos theta
divide both sides of the equation by cos theta, then
-98 TAN theta = -39.2
solve for Tan Theta, then arcTan Theta.
-98 TAN theta = -39.2
arc tan(-39.2)= -88.5386
-88.5386/-98 =.90345 would become the answer
I don't think I did this correctly...
No. I think you might need a tutor.
-98 TAN theta = -39.2
TAN theta= 39.2/98
Theta= arctan 39.2/98
To solve for theta, you can follow these steps:
1. Start with the equation: -98 TAN(theta) = -39.2
2. Divide both sides of the equation by -98: TAN(theta) = -39.2 / -98
3. Simplify the right side of the equation: TAN(theta) = 0.4
4. Take the inverse tangent (arctan) of both sides of the equation to isolate theta: theta = arctan(0.4)
5. Use a calculator or a table of values to find the arctan of 0.4: theta ≈ 21.8 degrees
So, the minimum angle at which the mass begins to slip is approximately 21.8 degrees.