Homework Help: Box On An Incline (Check Answer)

1. Sep 12, 2008

hp-p00nst3r

I have an answer for this question, but I am not quite sure if it's right so I would just like to run it by you guys and see what you think. The reason I am not sure is because I am not sure about how I reasoned that T should be 0. I used the reasoning that a rope can't exactly be "pushed".

1. The problem statement, all variables and given/known data
An 80kg man is trying to pull a 5 kg box up a 37° slope. The coefficient of static friction between the box and the ground is 0.45 and the coefficient of sliding friction is 0.40. The man is wearing shoes with a coefficient of static friction of 0.35 to the ground and coefficient sliding friction of 0.25 to the ground. Assuming the force the man pulls with is limited only by friction with the ground, how will the box accelerate at the instant shown?
http://img398.imageshack.us/img398/6179/220selftest81qw6.gif [Broken]

2. Relevant equations
F = ma

3. The attempt at a solution
http://img114.imageshack.us/img114/8350/mech220dynamics2ab5.jpg [Broken]

Last edited by a moderator: May 3, 2017
2. Sep 13, 2008

alphysicist

Hi hp-p00nst3r,

In your force equation for the man, you seem to be assuming that the man's acceleration is zero. The way I read the question, I think the man is moving up the ramp, at the same rate as the box.

3. Sep 14, 2008

hp-p00nst3r

Well I just tried it and I got 3.85m/s^2. The correct answer is supposed to be 2.78m/s^2.
http://img209.imageshack.us/img209/5355/mech220dynamics22es3.jpg [Broken]

Last edited by a moderator: May 3, 2017
4. Sep 14, 2008

alphysicist

When I read the problem, I skimmed over the word "trying" in "An 80kg man is trying to pull a 5 kg box up a 37° slope"; I though we could assume that he was actually succeeding at his task.

However, with the given value of the static friction of his shoes, he cannot even stand still on the ramp, much less pull up the box. So the man and box will be sliding down the ramp.

So here's how you can approach the problem. First pretend the rope is not there, and find out what accelerations the box and the man would have if they were not connected. You've already done this for the box and found out its acceleration would be 2.78 m/s^2 if the rope was not there.

Then do the same thing for the man: find his acceleration as if the rope was not there.

Now since the rope is there, the idea is that if the man's "natural" acceleration that you just found is less than the boxes (2.78m/s^2), then the man is dragging the box. In that case the rope would be taut and you would have to calculate the accelerations together.

If the man's natural acceleration is greater than the box's, then he is overtaking the box, and the rope is slack with zero tension.

Do you get the right answer?

5. Sep 14, 2008

hp-p00nst3r

I do get the right answer for this question with your method, but this method appears to only work for this question but it fails for the question after that. Your method does make logical sense to me so I am puzzled as to how it isn't working for the other question as well. The other question is essentially the same except that u_s = 0.6 and u_k = 0.5 for the man. Using the method you showed me, tension ends up being 0 again, which would make the acceleration of the box to be 2.78m/s^2, but the answer is 1.93m/s^2.

6. Sep 14, 2008

alphysicist

Looking at the numbers quickly, I do not believe the rope tension will be zero. What do you get for the acceleration of the man down the incline if the rope was not there?

7. Sep 14, 2008

hp-p00nst3r

Using the coefficient of kinetic friction for the man, I get 1.986m/s^2 down the slope and using the coefficient of static friction for the man, I get 1.20m/s^2 down the slope.

8. Sep 14, 2008

alphysicist

So if the rope wasn't there, the box would accelerate at 2.78m/s^2 and the man would accelerate at less than that, which would mean that the box would be leaving the man behind as it goes down the slope.

But since the rope is there, they have to move together, and so the man will slow down the box. This means the rope will have tension (because it is putting a force on both of them to hold them together).

9. Sep 14, 2008

hp-p00nst3r

Oh yeah, good point. I don't know what I was thinking back there. I redid the calculations now with tension included but now im getting 2.03m/s^2 down the slope
http://img139.imageshack.us/img139/2606/mech220dynamics23gg8.jpg [Broken]

Last edited by a moderator: May 3, 2017
10. Sep 14, 2008

alphysicist

If the only thing that changed was the frictional coefficients for the man's shoes, I don't see where you have done anything wrong. (Maybe check it again to be sure they did not slip a change past you?)

What confuses me is that you found that the acceleration of the man without the rope would be 1.98 m/s^2 down the incline; if you add the effect of the rope pulling downwards, that would increase the downward acceleration, not decrease it; so I don't see how it could be 1.93 m/s^2. Maybe I'm reading the problem wrong somehow.

11. Sep 15, 2008

hp-p00nst3r

Maybe this statement "Assuming the force the man pulls with is limited only by friction with the ground" is something significant here? I am not sure.

I overheard some other students asking for help on this exact same question. The teacher did mention the method you told me about involving calculating accelerations without the rope. However, he did not go into details about how the calculation was supposed to go. I will check with the teacher tomorrow. Thank you for your help!