so I'm doing the back of this worksheet and on the bottom there's a list of a whole bunch of formulas...
I'm asked to find the minimum coefficent of friction if the road was flat of a car which I still don't know how to do and posted a question about it below but anyways...
so I remember that u = tan (pheta) I don't remember how this is so I remember we derived it in class several months ago but don't remeber at least I think it was something like that were u = something something can you please show me this derrivation it had to something with trig laws and I forgot it
also I believe that it was u = tan (pheta) because on the back of my worksheet one formula wright next to Ff = u Fn is tan pheta = V^2/rg??????
were did this formula come from???
thanks
You need to throw the dang formulas away and learn physics. I suggest go through the exercise problems in your text. You are, in fact, lost.
To find the derivation of the formula u = tan(theta), which relates the coefficient of friction (u) and the angle of inclination (theta), we can start by considering a block on an inclined plane.
Let's assume that the block is on the verge of sliding down the plane. In this case, the horizontal component of the weight of the block (mg) is balanced by the force of friction (Ff), and the vertical component is balanced by the normal force (Fn).
To begin, we can decompose the weight (mg) into its components. The vertical component is given by mg * cos(theta), and the horizontal component is mg * sin(theta).
Now, let's consider the forces acting on the block. The force of friction (Ff) acts parallel and opposite to the horizontal component of the weight, while the normal force (Fn) acts perpendicular to the plane.
According to Newton's second law, the sum of the forces in the horizontal direction is equal to the mass of the object times its acceleration in that direction. Since the block is not sliding, the acceleration is zero. Therefore, the equation becomes:
Ff = mg * sin(theta)
The sum of the forces in the vertical direction is equal to the mass of the object times its acceleration due to gravity. The normal force (Fn) cancels out the vertical component of the weight, so the equation becomes:
Fn = mg * cos(theta)
Now, if we divide both equations, we get:
Ff / Fn = (mg * sin(theta)) / (mg * cos(theta))
Simplifying further:
Ff / Fn = tan(theta)
Finally, we can substitute the definition of the coefficient of friction (u = Ff / Fn) into the equation:
u = tan(theta)
That's how the formula u = tan(theta) is derived.
Moving on to the formula tan(theta) = V^2/rg, where V is the velocity of an object moving in a curved path of radius r, and g is the acceleration due to gravity:
This formula arises from considering a vehicle moving in a curved path. When the vehicle is on the brink of skidding, the centripetal force (Fc) acting towards the center of the curve is equal to the frictional force. Therefore:
Fc = Ff
Using the centripetal force equation, Fc = (mv^2)/r, where m is the mass of the object, v is the velocity, and r is the radius of the curve, and substituting Ff = uFn, we have:
(mv^2)/r = uFn
Since Fn = mg, we can rewrite the equation as:
(mv^2)/r = umg
Dividing both sides by mg, we get:
v^2 / r = ug
Finally, rearranging the equation, we have:
tan(theta) = V^2/rg
This formula shows the relationship between the angle of inclination (theta) and the velocity (V), radius (r), and acceleration due to gravity (g) of an object moving in a curved path.
I hope this explanation helps you understand the derivation of these formulas! Let me know if you have any further questions.