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Solving for the normal force (Physics assignment)

  1. Nov 26, 2008 #1
    Unfortunately, I'm totally lost on this question. I read the topic in the archive, and I know I can't ask for an answer to this, and frankly I don't want one. But is someone able to give me the process for doing this?

    1. The problem statement, all variables and given/known data

    A cart loaded with bricks has a total mass
    of 24.3 kg and is pulled at constant speed by
    a rope. The rope is inclined at 25.6 degrees above
    the horizontal and the cart moves 16.4 m on
    a horizontal floor. The coefficient of kinetic
    friction between ground and cart is 0.6 .
    The acceleration of gravity is 9.8 m/s^2 .
    What is the normal force exerted on the
    cart by the floor? Answer in units of N

    2. Relevant equations

    I'm not sure :confused: I'm assuming I would need FG=mg, but after that, I'm lost. It seems like no matter what equation I use, I'm missing a variable.

    3. The attempt at a solution

    Again, I apologize, but I have no clue how to begin solving for this. My guess is to solve for the force applied somehow, but I have no idea how to.
     
  2. jcsd
  3. Nov 26, 2008 #2

    PhanthomJay

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    Draw a Free Body Diagram of the cart, and note ALL the forces acting on it. The gravity force (its weight) is just one of them. You also have a friction force, a normal force, and that applied rope tension force. Use Newton's laws in the x and y direction to solve for the unknown forces. Note that the problem states that the cart is moving at constant speed, which implies no acceleration.
     
  4. Nov 26, 2008 #3
    Before I start, thanks for the quick response.

    Ok, sorry about being unclear in the first post - I knew there was a force of friction/applied force, I just didn't say so.

    I was thinking about F=ma, but since acceleration=0, wouldn't that create a force of 0, meaning no friction, and when solving for the normal force using FF=[tex]\mu[/tex]FN, the answer for the normal force would be zero.

    I apologize, I'm sure I'm just thinking of this in a way that is wrong, but I find once I get the thought in my head it's difficult to ignore it.
     
  5. Nov 26, 2008 #4

    PhanthomJay

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    OK, and the normal force.
    that implies a net force of zero in both the x and y directions. The total forces or components of forces in the x direction, and in the y direction, must algebraically sum to 0.
    but you have noted already that there is friction
    Well let's clear your head and give it another shot. Look in the x and y directions separately. You'll get 2 equations with 2 unknowns, which you can solve.
     
  6. Nov 26, 2008 #5
    >_> Not quite sure why I didn't mention the normal force in my second post.

    Anyways, thinking about the X direction, I could assume that FF will be negative, and FAX will be positive. But if I were to use Fnet=FF+FAX, I'm still left with two variables to solve for.

    For the Y direction, Fnet=Fgravity+FN+FAY. Fgravity works out to -238.383 (negative due to direction), but I'm still left with 2 variables to solve for.

    Sorry for being impatient, but I have a horrible habit of leaving things until the last minute - this is due at 12:00AM tonight (it's 9:05 in my timezone). I understand that you're being the best teacher possible (and I don't say that because I'm trying to suck-up to you) by having me figure out the problem on my own, but if it gets to the point where you have to log off for the night, and I can't figure it out, would you be able to give me a formula to use, or a set of steps to follow to solve this? I really appreciate the help you've given me, but if you had to log off, I still don't know that I would be able to finish this.

    Edit - It appears I'm too late to talk about you logging off :S
     
    Last edited: Nov 26, 2008
  7. Nov 26, 2008 #6

    PhanthomJay

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    that's right, and that's why you need another equation. Let's designate the applied rope force as T (too many F's spoil the pot). So far you have Tcos25.6 - uN = 0.
    YES, so its Tsin25.6 + N - 238.4 = 0
    You've got everything you need now. Just solve those 2 equations for the 2 unknowns T and N.
     
  8. Nov 26, 2008 #7
    (Again, I'm sorry for being so totally clueless. I did fine in physics up until this unit.)

    Tcos25.6-uN = 0 and Tsin25.6 + N - 238.4 = 0 sounds like a setup for a 2x2 elimination question that I did in grade 10 math. But is that the solution to my problem? (If you don't understand what I mean, I'm talking about subtracting one equation from the other, after multiplying to make the absolute value of one variable the same, with one negative and one positive. Although I'm pretty sure that sentence made it even more confusing.) I've never used elimination in physics yet, but it seems to be the only way to solve this.

    Edit - By the way, I forgot to mention this, but I can't thank you enough for the all the help.

    2nd edit - I'm assuming that u=coefficient of friction. Correct me if I'm wrong please.

    3rd edit - Ugh. I worked out the elimination, submitted my answer to the online-assignment page...and it's wrong. Is there any more that you can do to help? I hate asking you to hand it to me, but I also don't want to receive a crappy grade on this assignment.
     
    Last edited: Nov 26, 2008
  9. Nov 26, 2008 #8

    PhanthomJay

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    Yes, your assumptions are all correct. Get out your grad 10 math book and solve those 2 equations by any means you are familiar with. The clock is ticking... are you from Bermuda?
     
  10. Nov 26, 2008 #9
    Nope, I'm in Nova Scotia, Canada. Same time zone though. Anyways, I'll copy over my last edit, in case you didn't read it (I edited after you had posted a reply)

    3rd edit - Ugh. I worked out the elimination, submitted my answer to the online-assignment page...and it's wrong. Is there any more that you can do to help? I hate asking you to hand it to me, but I also don't want to receive a crappy grade on this assignment.

    My elimination worked out to 223N for T, and 142 for N (I assumed this was the force normal, if not, what was it?) Was there a problem with my math?
     
  11. Nov 26, 2008 #10

    PhanthomJay

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    OK, looks like

    (1) T(.901) - 0.6N = 0
    (2) T(.432) + N = 238.1

    Multiply the 2nd equation by .6 to get

    (3) T(.26) + .6N = 142.9

    Now add equations (1) and (3) to get

    (4) 1.16T = 142.9

    From which, T = 123.2. But the problem wants you to find N, the normal force. This one's on you.
     
  12. Nov 26, 2008 #11
    There's no way I can tell you how much I appreciate you helping me out, but still, thank you so much.
     
  13. Nov 26, 2008 #12

    PhanthomJay

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    You're welcome. As I look out my window facing east, I see the Atlantic Ocean (I live on the east cost, mainland ,USA). I was trying to figure who was one hour ahead of me in time, and I thought about Bermuda. Forgot all about Nova Scotia. Stay warm!
     
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