Block on an inclined plane that is accelerating sideways

After plugging that in for the lower value I got a = 0. So would one of the angles be arctan(μ), being the smallest angle that the block can remain stationary on the rightward accelerating plane?And then the other arccot(μ), being the angle at which the block can no longer remain on the plane?In summary, the block of mass m on an inclined plane with angle o and acceleration a to the right will remain at rest for a range of values of a. The lowest value of a is given by g((sin(o) - ucos(o))/(cos(o) + usin(o))), while the highest value is given by g((sin(o) + ucos(o))/(
  • #1
FScheuer

Homework Statement


[/B]
A block of mass m rests on an inclined plane with angle o accelerating to the right with acceleration a. The coefficient of static friction is u.

Homework Equations



For what range of a does the block remain at rest on the plane.

The Attempt at a Solution


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I started by drawing a free body diagram with a gravitational force, normal force, and frictional force. The horizontal component of my frictional force is in the direction of the acceleration. I believe solving for this should give me the lowest value of a where the block remains at rest. I set all of the horizontal comments equal to ma (Fncos(o) + uFncos(o) = ma), and all of the vertical components to 0 (Fnsin(o) - mg - uFnsin(o) = 0). I then set each of those equations equal to Fn, the normal force, and then set them equal to each other (ma/(cos(o)(1 + u)) = mg/(sin(o)(1 - u)). For the lowest value of a, I got a = gcot(o).
Is what I’ve done so far correct?
 
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  • #2
Not correct. First, make sure the question is correct. The block is either at rest or accelerating. Can't be both.
FScheuer said:
For what range of a does the block remain at rest on the plane.
If this is the real question, the answer is not a range, but a very specific value. Hint: it is not non-zero.
 
  • #3
lewando said:
Not correct. First, make sure the question is correct. The block is either at rest or accelerating. Can't be both.

If this is the real question, the answer is not a range, but a very specific value. Hint: it is not non-zero.
The block is at rest with respect to the plane for a range of values of a.

Here is a picture of the question.
 

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  • #4
FScheuer said:
Here is a picture of the question.
Can you Upload a picture of the problem (see the button in the lower right)? And Upload your work so far, including your FBD. Thanks.
 
  • #5
berkeman said:
Can you Upload a picture of the problem (see the button in the lower right)? And Upload your work so far, including your FBD. Thanks.

I looked over my work and got a slightly different answer. I now have gcot(o)((1 + u)/(1 - u))

Anyway, here is my work so far and the question. I would think that the work that I did would give me the lowest value of a in the range. Then to get the highest value I would do the same thing expect with the frictional force acting in the opposite direction.
 

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  • #6
FScheuer said:
I looked over my work and got a slightly different answer. I now have gcot(o)((1 + u)/(1 - u))

Anyway, here is my work so far and the question. I would think that the work that I did would give me the lowest value of a in the range. Then to get the highest value I would do the same thing expect with the frictional force acting in the opposite direction.
Check your uses of sine and cosine. I find it helps to consider extreme cases. E.g. if θ=0, what would the horizontal component of the normal force be? With experience, you might also recognise that equations involving friction generally have terms FN and FNμ with different trig functions, not the same one.
 
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  • #7
haruspex said:
Check your uses of sine and cosine. I find it helps to consider extreme cases. E.g. if θ=0, what would the horizontal component of the normal force be? With experience, you might also recognise that equations involving friction generally have terms FN and FNμ with different trig functions, not the same one.
I switched the cos and sin on the normal force. I had those mixed up. Now my answer for the lowest value of a is

a = g((sin(o) + ucos(o))/(cos(o) - usin(o)))
 
  • #8
FScheuer said:
I switched the cos and sin on the normal force. I had those mixed up. Now my answer for the lowest value of a is

a = g((sin(o) + ucos(o))/(cos(o) - usin(o)))
That is certainly one critical value, but how do you conclude it is the lowest of the range?

By the way, writing tan(α) for μ yields interesting forms of the equations.
 
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  • #9
haruspex said:
That is certainly one critical value, but how do you conclude it is the lowest of the range?

By the way, writing tan(α) for μ yields interesting forms of the equations.
My bad, I now think it should be the highest possible value because I chose the frictional force to be acting down the incline. If I do the same thing with the frictional force acting up the incline, I think I will get the lowest possible value.
 
  • #10
FScheuer said:
My bad, I now think it should be the highest possible value because I chose the frictional force to be acting down the incline. If I do the same thing with the frictional force acting up the incline, I think I will get the lowest possible value.
Agreed.
 
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  • #11
I now have a = g((sin(o) - ucos(o))/(cos(o) + usin(o))) for the low value, and a = g((sin(o) + ucos(o))/(cos(o) - usin(o))) for the high value. I’m not quite sure what they want when they ask for the special angles. Is it just the angle where the lowest value is equal to 0 and the angle where there is a vertical asymptote?
 
  • #12
FScheuer said:
when they ask for the special angles. Is it just the angle where the lowest value is equal to 0 and the angle where there is a vertical asymptote?
I would say so.
Did you try writing tan(α) for μ as I suggested?
 
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  • #13
haruspex said:
I would say so.
Did you try writing tan(α) for μ as I suggested?
After plugging that in for the lower value I got a = 0.

So would one of the angles be arctan(μ), being the smallest angle that the block can remain stationary on the rightward accelerating plane?

And then the other arccot(μ), being the angle at which the block can no longer remain on the plane?
 
  • #14
FScheuer said:
arctan(μ), being the smallest angle that the block can remain stationary on the rightward accelerating plane?
No. The equation in this case is for the least acceleration consistent with staying put, and that value of the angle says that that least acceleration is zero.
FScheuer said:
arccot(μ), being the angle at which the block can no longer remain on the plane?
No. The equation in this case is for the largest acceleration at which it will stay put. The value arccot(μ) gives an infinite acceleration. So with that angle, at what acceleration would the mass slip?
 
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  • #15
haruspex said:
No. The equation in this case is for the largest acceleration at which it will stay put. The value arccot(μ) gives an infinite acceleration. So with that angle, at what acceleration would the mass slip?
The block shouldn’t slip regardless of the acceleration at that point.

So arctan(μ) and arccot(μ) should be the special angles, the former being the angle where the lowest value of a is 0, and the latter being where there is no value of a too large for the block to remain stationary relative to the plane.
 
  • #16
FScheuer said:
The block shouldn’t slip regardless of the acceleration at that point.

So arctan(μ) and arccot(μ) should be the special angles, the former being the angle where the lowest value of a is 0, and the latter being where there is no value of a too large for the block to remain stationary relative to the plane.
Right.
But something is bothering me... what if μ>1?
 
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  • #17
haruspex said:
Right.
But something is bothering me... what if μ>1?
I can’t think of any reasons that there would be problems or anything special when μ > 1. What exactly do you mean by that?
 
  • #18
FScheuer said:
I can’t think of any reasons that there would be problems or anything special when μ > 1. What exactly do you mean by that?
It was not specifically μ>1, but considering large μ made me realize we have something not quite right in the first part of the question.
Writing tan(α) for μ your min and max expressions become ##\tan(\theta-\alpha)\leq\frac ag\leq\tan(\theta+\alpha)##.
What if ##\theta+\alpha>\frac \pi 2##?##
 
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  • #19
haruspex said:
It was not specifically μ>1, but considering large μ made me realize we have something not quite right in the first part of the question.
Writing tan(α) for μ your min and max expressions become ##\tan(\theta-\alpha)\leq\frac ag\leq\tan(\theta+\alpha)##.
What if ##\theta+\alpha>\frac \pi 2##?##
I’m not quite sure where the a is coming from. The interesting values seem to come when I plug in tan(theta) and cot(theta) in for μ rather than tan(a).

I think I might see what you’re talking about though. The highest value of a where the block can remain at rest approaches infinity, but then if the angle is large enough, that value of a becomes negative and lies to the right of the asymptote. I’m not quite sure why this is happening. Is my second equation wrong?
 
  • #20
FScheuer said:
I’m not quite sure where the a is coming from.
Not a, α (alpha). Just choosing to represent μ in terms of an angle, because of the way it simplifies the equations.
FScheuer said:
Is my second equation wrong?
It is incomplete, in the sense that the equation is only valid over a certain range.
Can you tell me what is wrong with this statement:
"If a body is not sliding on a surface with which it is in contact under a normal force N then the frictional force is μsN"
?
 
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  • #21
haruspex said:
Not a, α (alpha). Just choosing to represent μ in terms of an angle, because of the way it simplifies the equations.

It is incomplete, in the sense that the equation is only valid over a certain range.
Can you tell me what is wrong with this statement:
"If a body is not sliding on a surface with which it is in contact under a normal force N then the frictional force is μsN"
?
The frictional force is not necessarily equal to that value, but can resist a force that is less than or equal to it.
 
  • #22
FScheuer said:
The frictional force is not necessarily equal to that value, but can resist a force that is less than or equal to it.
Right. So how would you qualify the equation?
 
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  • #23
haruspex said:
Right. So how would you qualify the equation?
For the equation that gives the highest value of a, would it just be theta < arccot(u)?
 
  • #24
FScheuer said:
For the equation that gives the highest value of a, would it just be theta < arccot(u)?
You need to break it into cases. If cot(θ)>μ then ... else ...
 
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  • #25
haruspex said:
You need to break it into cases. If cot(θ)>μ then ... else ...
Would the equation for the high value of a work for all angles between 0 and arccot(u) and then all angles between arccot(u) and 90 have no maximum value of a for which the block doesn’t rest on the plane?
 
  • #26
FScheuer said:
Would the equation for the high value of a work for all angles between 0 and arccot(u) and then all angles between arccot(u) and 90 have no maximum value of a for which the block doesn’t rest on the plane?
Yes.
 
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  • #27
haruspex said:
Yes.
I understand conceptually why this is the case, but can you explain mathmatecally why the equation I found works for some value but not others, and how to figure out which values work?
 
  • #28
FScheuer said:
I understand conceptually why this is the case, but can you explain mathmatecally why the equation I found works for some value but not others, and how to figure out which values work?
Your derivation assumed the frictional force was μsN. That is valid as long as the acceleration can be made so large that such a high frictional force is needed. When that is no longer true the equation is incorrect.
 
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Related to Block on an inclined plane that is accelerating sideways

1. What is the relationship between the angle of inclination and the acceleration of the block?

The angle of inclination has a direct impact on the acceleration of the block. As the angle increases, the acceleration also increases in the same direction as the angle. This is due to the gravitational force being resolved into two components, one parallel to the surface and one perpendicular. The parallel component is responsible for the acceleration of the block.

2. How does the mass of the block affect its acceleration on an inclined plane?

The mass of the block does not affect its acceleration on an inclined plane, as long as the surface is frictionless. This is because the gravitational force acting on the block is directly proportional to its mass, so the mass cancels out in the equation for acceleration. However, if there is friction present, the mass will have an impact on the acceleration as it affects the magnitude of the frictional force.

3. Can the block's acceleration ever be greater than the acceleration due to gravity?

No, the block's acceleration can never be greater than the acceleration due to gravity. This is because the force of gravity is the only force acting on the block, and it is always directed downwards. Any acceleration of the block will be in the same direction as the force of gravity, and thus, will be limited by the acceleration due to gravity.

4. How does the coefficient of kinetic friction affect the acceleration of the block on an inclined plane?

The coefficient of kinetic friction has a significant impact on the acceleration of the block on an inclined plane. It is directly proportional to the frictional force, which acts in the opposite direction of the block's motion. As the coefficient of kinetic friction increases, the frictional force also increases, causing the block's acceleration to decrease.

5. What happens to the block's acceleration if the angle of inclination is 90 degrees?

If the angle of inclination is 90 degrees, the block will not accelerate at all. This is because the gravitational force is acting perpendicular to the surface, and there is no parallel component to cause acceleration. The block will simply remain at rest or continue to move at a constant speed in the direction of the incline.

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