Confusion over an Inclined Plane Question

[Tyler_Cavan]

Homework Statement

A box of textbooks of mass 25.0 kg rests on a loading ramp that makes an angle "a" with the horizontal. The coefficient of kinetic friction (muk) is 0.230 and the coefficient of static friction (mus) is 0.340.
a) As the angle "a" is increased, find the minimum angle at which the box starts to slip.
b) At this angle, find the acceleration once the box has begun to move.
c) At this angle, how fast will the box be moving after it has slid a distance 5.30 along the loading ramp?

2. Homework Equations /Attempts

I have solved a), with an answer of 19 degrees, and my answer for b) does not seem to be right although I can't see where I went wrong. I cannot complete c) till b) is finished.

for b):

F= ma
mgsin"a" - (muk)mgsincos"a" = ma
therefore:
a = g*sin"a" - (muk)*g*cos"a"
and I get a = 1.059 which is wrong.

Any insight is appreciated.

 

[Doc Al]

for b):

F= ma
mgsin"a" - (muk)mgsincos"a" = ma
therefore:
a = g*sin"a" - (muk)*g*cos"a"

Good. (I assume the "sincos" was a typo.)

and I get a = 1.059 which is wrong.

Try using a more accurate value for the angle.

 

[Tyler_Cavan]

- Thank you. sincos was an error. It was meant to be just cos.

The questions asks me to express myself to two significant digits, and when I do whether I input 19 degrees or 18.77803322 degrees, I get the same, "1.0". Do you know what it could be?

Thanks, Ty.

 

[Doc Al]

Beats me. Your answer is correct.

 

[Tyler_Cavan]

Bugger of a question, ain't it?

Is there anyway I can do c) without using acceleration? If I do, I am going to assume that a = 1.0 since I, myself, also can't see anything wrong with it. But you'd know better. What equation can I then use.

Thanks, Ty.

 

[Doc Al]

You can use energy methods, but they end up using the same values as you calculated for acceleration--so don't expect anything different.

I would use the more accurate value for acceleration and just round off your answer at the end.

(Unfortunately, these online systems can be flaky. Don't think this is the first time that such puzzlers have been reported.)

 

[Tyler_Cavan]

That's why I like the good old fashioned "write-it out" technique.

If I was going to find d then, can I use:
vf^2 = vi^2 +2ad
or would it be wrong to assume I started from rest?

 

[Doc Al]

If I was going to find d then, can I use:
vf^2 = vi^2 +2ad

Yes.

or would it be wrong to assume I started from rest?

That would not be wrong--assume it starts from rest.

 

[Tyler_Cavan]

Thank you, Doc Al. Weirdly enough it accepted my answer for d, even though I used 1.0 as the acceleration.

Would you mind if I asked you another question?

 

[Doc Al]

Ask away.

 

[Tyler_Cavan]

Thank you again.

< http://people.hws.edu/faux/Phys1/practice2.pdf >

Question number 5:
5. The 4.00 kg block in the figure to the right (see link) is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are
extended as shown in the diagram and the tension in the upper string is 80.0 N.
a) What is the tension in the lower cord?
b) How many revolutions per minute does the system make?
c) Find the number of revolutions per minute at which the lower cord just goes slack.


I have T figured out to 30, but I am stuck on finding b) and c). I know I need to find a, to get v to get T but I don't know how to get a.

I have tried:
F=ma
79 - 30 - 39.2 = 4*a
a=2.45
then, a = v^2/R
v= 1.355 544 171
then, T = (2*pi*R)/v
T= 3.4576 381 723 rev/s
= 208.582 903 4 rev/min
which I know is too high. What am I doing wrong?

When I do c)
F=ma
79 - 0 - 39.2 = 4*a
a=9.95
then, a = v^2/R
v= 2.731 757 676
then, T = (2*pi*R)/v
T= 1.725 039 165 rev/s
= 103.502 349 9 rev/min
This seems to low compared to b).

By the way, I have used R=0.75. I did calculate that right, right?

 

[Doc Al]

I have T figured out to 30, but I am stuck on finding b) and c). I know I need to find a, to get v to get T but I don't know how to get a.

The acceleration you need is the centripetal acceleration, which in this case is horizontal. What horizontal forces act on the mass?

I have tried:
F=ma
79 - 30 - 39.2 = 4*a

Forces are vectors--direction counts. Treat horizontal and vertical components separately. Once you straighten this out you can redo parts b and c.

By the way, I have used R=0.75. I did calculate that right, right?

Right.

 

[Tyler_Cavan]

The acceleration you need is the centripetal acceleration, which in this case is horizontal. What horizontal forces act on the mass?

To find centripetal acceleration is a = v^2/r, but I do not have v. I do know that v and a are at perpendicular to one another though. Right?
I know centripetal force (F= (mv^2)/R) is a horizontal force acting on the mass. By the way, what about centrifugal "force"; does it play a role?


I thought my signs where correct; may I ask how I was mistaken? No insult meant, but you probably right. As for the treating horizontal and vertical components separately, how can I set it up? I don't know what I am going to resolving for, but I am going to play around with it now.

 

[Doc Al]

To find centripetal acceleration is a = v^2/r, but I do not have v. I do know that v and a are at perpendicular to one another though. Right?
I know centripetal force (F= (mv^2)/R) is a horizontal force acting on the mass.

Since the mass is whirling in a circle, you know it's centripetally accelerating. What forces are producing the centripetal acceleration? (That's why I asked: What horizontal forces act on the mass?)

Once you find the net force towards the center, you'll set that equal to the formula for centripetal force.

By the way, what about centrifugal "force"; does it play a role?

No. Centrifugal force is a "fictitious" force that arises when analyzing things from a rotating reference frame. Not relevant here. (You could use it to solve the problem. Don't.)

FYI: "Centripetal" means "towards the center" and "centrifugal" means "away from the center".

I thought my signs where correct; may I ask how I was mistaken?

The problem is not just signs. There are three forces acting on the mass: the two tensions and gravity. Find the net horizontal force acting on the mass. (Add up the horizontal components of each force.)

 

[Tyler_Cavan]

Once you find the net force towards the center, you'll set that equal to the formula for centripetal force.

I understand now! I took what you told me, about my signs and horizontal forces, and I got my answer of 44.5 revolutions per minute, and then for part c) I and got 29.9 rev/ min. I appreciate all the help you gave me, man. Thanks, Ty.