Will the Block Slide on an Accelerating Inclined Plane?

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The discussion centers on a physics problem involving a block on an accelerating inclined plane with friction. The key equation derived suggests that for the block to slide, the acceleration must exceed a specific threshold, expressed as a > g tan(Theta - Alpha). Participants clarify that static friction allows the block to maintain the same acceleration as the incline, preventing it from sliding until that threshold is surpassed. The concept of static friction is further explored, emphasizing its role in enabling the block to "keep up" with the incline rather than acting as a stopping force. Overall, the conversation effectively addresses the mechanics of the problem and the role of static friction in motion.
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Homework Statement


I have an incline with a degree of Alpha and it is with friction that is equal to (Us = tan theta)
It is accelerating to the right. Prove that this equation if it is true then the block will slide.
a > g tan(Theta - Alpha)

The Attempt at a Solution


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That was my thought of forces.So I figured out that if I want them to stick together then it should have the same acceleration and orientation.
So the Net of forces should be ma and has the same orientation as (a)
I have encountered a problem before with 2 stacked objects and found that the static force keeps the blocks together by giving it the same acceleration ( Correct me if I am wrong, I kinda get confused on how that happens) and also I can't find any force that could be acting that way so it produces the same acceleration and orientation. Anyway as a result, I came up with these equations.

Friction - mg sin alpha = ma cos alpha (1)
mg cos alpha - N = ma sin alpha (2)
Friction = Us N
Then by putting 2 in 1
Us (mg cos alpha - ma sin alpha) - mg sin alpha = ma cos alpha
Us as mentioned is equal to tan theta
Tan theta mg cos alpha - tan theta ma sin alpha - mg sin alpha = ma cos alpha
Divide all of it by cos alpha and m
Tan theta g - tan theta a tan alpha - g tan alpha = a
Do some math and you will come up with this
g(tan theta - g tan alpha) = a ( 1 + tan theta tan alpha)
a = g (tan theta - g tan alpha) / ( 1 + tan theta tan alpha)
a = g tan( theta - alpha)
If I want it to slide then a bit more acceleration will make it slide then a > g tan(tehta - alpha)

I highly doubt that the solution is right but I hope so. (Ehhhh worth trying right??) My conflict is with the static friction how it makes things move because that what I understood from this problem.
http://www.uwgb.edu/fenclh/problems/dynamics/multi/1/
He put a value of f and saw if that value of f equals to the static friction or no. I hope someone clear this misconception for me.

Thanks. [/B]
 
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Hello Biker, :welcome:

I think you are doing fine. You are given a static friction coefficient ##\mu_s## and that gives you the maximum friction force. With that you calculate the maximum acceleration of the incline for which the block can 'keep up'. Any more and it will start to slide. And that's all that is asked of you in this exercise.

Well done !
 
BvU said:
Hello Biker, :welcome:

I think you are doing fine. You are given a static friction coefficient ##\mu_s## and that gives you the maximum friction force. With that you calculate the maximum acceleration of the incline for which the block can 'keep up'. Any more and it will start to slide. And that's all that is asked of you in this exercise.

Well done !

You just made my day, I have been working on this problem for so long. Although none of these are in my books or exams. So I have a question on a stacked objects problems or any other problem as this, The static friction force is what keeps the body moving with it but sometimes I find it hard to understand how the static force is acting on giving acceleration not trying to stop a object but If I think about it in another way If I consider that the block is moving backwards relatively to the incline then it is rational to consider that the static friction is acting on it to make it go with the acceleration same as the incline, Is that right?
 
Yes (to answer your last question)

I prefer the term 'keeping up' to the expression 'moving backwards' in this situation: the whole thing is that the block accelerates with the same acceleration as the incline: it does not move with respect to the incline. That's why only the coefficient of static friction ##\mu_s## is enough to answer the question.
 
Hmm, So In "Keeping up" terms the static friction work here not as a stopping force it makes the block keep up with the incline. It is not fully understood to me but I will try to think about it more.

Thanks for helping.
 
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