More friction problems!

Impathy · Sep 21, 2005

Hello ... looking for some assistance! I have one two-part question I'm struggling with. I got the first part of the first question:

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 160 N at an angle of 35 degrees above the horizontal. The box has a mass of 25.8 kg and the coefficient of friction between box and floor is 0.254. Find the acceleration.

Alright, so here is what I did: F(N)+160*sin(35)-mg=0, F(N)=161.06777 N
F(f)=uN=0.254-161.06777=40.9112
F=160*cos(35)-40.9112=90.15311=ma
a=91.15311/25.8=3.4943 m/(s^2)
And that's correct ... now the second part is just that the student starts moving the box up a 13.2 degree incline, keeping the same 160 N force directed at 35 degrees.

Here is what I tried first:
F=160*cos(35)-mgsin(13.2)-40.9112=32.416=ma
a=32.416/25.8=1.256 m/(s^2) ... WRONG

So then I thought that since I rotated the diagram by 13.2 degrees, it should have affected the 35 degrees:
F(N)+160*sin(35-13.2)-mg=0, F(N)=-193.42 ... this doesn't seem right ... try it anyway ...
F(f)=uN=0.254--193.42=-49.1289
F=160*cos(35-13.2)-mgsin(13.2)+49.1289=106.696=ma
a=106.696/25.8=4.1355 m/(s^2) ... WRONG AGAIN

I've drawn my free body diagrams over and over and I just don't see what to do in the second part where it's inclined 13.2 degrees. I know must be setting it up all incorrectly. Any hints, please? Thanks in advance!

JoAuSc · Sep 21, 2005

First, it would help if you got an algebraic solution to your problem first, then substituted the numbers. For example, for the first part of the problem I solved it algebraically to get:

a = ( F_a*cos(theta) - umg )/m

(F_a is the force of the pull, the applied force.)

Anyways, you computed the acceleration in the horizontal direction. However, a box moving up an incline is not moving horizontally. The easiest solution would be to rotate your axes by 13.2 degrees so that the x-axis once again lines up with the path of the box. Another thing you could do is keep the axes the way they are and find the acceleration in the vertical direction, then use the pythagorean theorem to find the magnitude of the acceleration.

Impathy · Sep 21, 2005

Alright, I see what you mean. I guess I'm confused, because I thought I was rotating the axis already (hence the mgsin). I tried just solving it algabraeically like you suggested ... when I rotate my axis, would my equation look like this:

[tex] a = \frac{F cos \theta - mg sin \phi - \mu mg}{m}[/tex]

or this:

[tex] a = \frac{F cos \left(\theta-\phi\right) - mg sin \phi - \mu mg}{m}[/tex]

with [tex]\theta = 35^o[/tex] and [tex]\phi = 13.2^o[/tex] ??

Or do I have both of these all messed up? I'll try the Pythagorean theorem now, though ... Thanks!

JoAuSc · Sep 21, 2005

Impathy said:

Alright, I see what you mean. I guess I'm confused, because I thought I was rotating the axis already (hence the mgsin). I tried just solving it algabraeically like you suggested ... when I rotate my axis, would my equation look like this:

[tex] a = \frac{F cos \theta - mg sin \phi - \mu mg}{m}[/tex]

or this:

[tex] a = \frac{F cos \left(\theta-\phi\right) - mg sin \phi - \mu mg}{m}[/tex]

with [tex]\theta = 35^o[/tex] and [tex]\phi = 13.2^o[/tex] ??

Or do I have both of these all messed up? I'll try the Pythagorean theorem now, though ... Thanks!

The second one seems closer to being correct, because while the absolute direction of the applied force wouldn't change, its angle relative to your new set of axes would. The only thing that worries me is the friction term ([tex]\mu mg[/tex]). You assume that the force of the floor on the box is the same even though the floor is at an incline. What you need to do is calculate the forces perpendicular to the incline (which should sum to zero since the box isn't accelerating in that direction), solve for an expression for the normal force, then use it instead of "mg" for the friction force.

Fermat · Sep 21, 2005

Is the answer a = 1.322 m/s² ?

Impathy · Sep 21, 2005

Fermat said:

Is the answer a = 1.394 m/s² ?

I don't know the answer ... I get one more chance to submit, so I want to be sure! And I want to understand it ... how did you come about that answer? Did you use a formula similar to one above?

Impathy · Sep 21, 2005

JoAuSc said:

The second one seems closer to being correct, because while the absolute direction of the applied force wouldn't change, its angle relative to your new set of axes would. The only thing that worries me is the friction term ([tex]\mu mg[/tex]). You assume that the force of the floor on the box is the same even though the floor is at an incline. What you need to do is calculate the forces perpendicular to the incline (which should sum to zero since the box isn't accelerating in that direction), solve for an expression for the normal force, then use it instead of "mg" for the friction force.

Ahh, I see what you mean. Does this seem reasonable then?
[tex] a = \frac{F cos \left(\theta-\phi\right) - mg sin \phi - \mu mgcos\phi}{m}[/tex]

... Wait, I think I get part of what you're saying ... maybe not all of it. Hm.

If I use the above equation I get 2.9518 m/s², which seems physically reasonable compared to the 3.49 m/s² on a flat surface, doesn't it?

Fermat · Sep 21, 2005

The books are being pulled along a plane, initially horizontal.
When the plane is titlted to an angle of 13.2 degrees, then you have an additioanal component of Mgsin@ down the plane opposing the component of the tension along the plane, Tcos(35). You also habve Mgcos@ as the new normal component of the mass of the books. The new normal reaction is that much lees than before. N=Tsin35 - Mgsin@. (@ = 13.2 degrees)
Recalculate N. Recalculate the accelerating force. do the F=ma bit

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