Basics: Object on Incline, Forces (Im Hopeless) Help

AI Thread Summary
An object weighing 100 grams on a 21-degree incline begins to slide when it overcomes an unknown coefficient of friction. At this point, the object does not accelerate, indicating a constant velocity, as the forces acting on it are balanced. The x-component of the forces is correctly identified as mgsin(theta) - mumgcos(theta), while the y-component should account for the normal force and gravitational force perpendicular to the incline. To find the coefficient of friction, it is established that it can be calculated using the equation 0 = mgsin(theta) - mumgcos(theta), leading to mu = tan(theta). The discussion emphasizes that the coefficient of friction is dependent on the angle at which the object begins to slide.
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Homework Statement



Object weighing 100 grams is on a surface which is inclined to an angle where the mass just overcame the coefficient of friction (which is unknown) and started sliding down. Angle was measured to be 21 degrees to the horizontal.

What I want to know is would the mass have a constant velocity, thus acceleration of 0?
Why?

Also is this right in resolving the problem into x and y components:
x-component = mgsintheta - mumgcostheta (where mu = coefficient of friction)
y-component = None, as the object does not move in directions perpendicular to the plane.


Would greatly appreciate help, stressed out and struggling all day...
 
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yes, the block is first at rest so v=0, and when it JUST slips and starts to slide down it isn't accelerating because the angle is measured as low as possible so that the object just overcomes the coefficient of friction.

your x component is right but for your y you should have N-mgcostheta or mgcostheta-N depending on how you setup your coordinates.
 
Ok thank you very much !

Now is it possible to calculate the value of the coefficient of friction without using tan theta?
Resolving the situation into x and y components, and assuming gravity to be 9.8m/s^2,

m*a = (mgsintheta) - (mumgcostheta)
m = 0.1kg
a = 0
theta = 21 degrees

0 = 0.3512 - (mu * 0.9114)

How do I go from here?

Sorry to be a pain...
 
well if you solve 0=mgsintheta-mumgcostheta you get mu=tantheta.

You can also continue it numerically how you have it. just solve for mu in 0=.3512-.9114mu which is the same as taking tan theta.

I don't think there's a way to solve for mu without dealing with angle's since the coefficient of friction depends on the angle at which the object begins to slip or slides at a constant velocity.
 
http://www.pha.jhu.edu/~broholm/l10/node2.html

If you use LaTeX the text will be more legible. It's easy, and it's fun.

Some remarks...

If you choose x to be parallel and y perpendicular to the incline:

F_x = mg \sin \theta - \mu F_N

F_y = F_N - mg \cos \theta

If no motion in the y direction F_y = 0

If the coefficient of friction is the same whether the object is at rest or in motion there will be no acceleration and F_x = 0.

But if the friction is lower when the object is in motion there will be an acceleration.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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