1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A box weighing 229 N at 35°

  1. Oct 29, 2006 #1
    A box of books weighing 229 N is shoved across the floor by a force of 455 N exerted downward at an angle of 35° below the horizontal.

    (a) If µk between the box and the floor is 0.57, how long does it take to move the box 5 m, starting from rest?

    (b) If µk between the box and the floor is 0.75, how long does it take to move the box 5 m, starting from rest?

    I started by finding the x and y vectors x=272.71 y=260.98 Then i tried to th Neutral force and used it in the equation Fk=µk*N But i am lost and dont know hwat to do please help me
  2. jcsd
  3. Oct 29, 2006 #2
    Draw a free body diagram. Remember that N is not simply mg, but also the magnitude of the y-component of the acting force.


    You know mg (229 N), [itex]F_{push}[/itex] (455 N), and [itex]\theta[/itex] (35[itex]^{\circ}[/itex] SE).
    Last edited: Oct 30, 2006
  4. Oct 30, 2006 #3
    i found acceleration 15.74 and then found time to be 1.694s is this correct?
  5. Oct 30, 2006 #4
    I haven't solved the problem, so I don't know if that answer is correct. However, you should ask yourself, does an acceleration of almost 16 [itex]m/s^{2}[/itex] seem reasonable? That's over 30 mph/s, and the box is moving 5 meters in under 2 seconds. I'd say it's probably not correct.

    Show your work, and we can pick out the error.
    Last edited: Oct 30, 2006
  6. Oct 30, 2006 #5
    Actually, it sounds reasonable. Look at the forces involved on a not so massive item.

    EDIT: cos(35) * 455N * 9.8m/s^2 / 229N = 15.95
    Last edited: Oct 30, 2006
  7. Oct 30, 2006 #6
    You're saying it's reasonable for a force of 445 N (roughly equal in magnitude to the weight of a pre-teen girl)[itex]-[/itex]and not the full force, but a component of it[itex]-[/itex]to accelerate a 50 pound object to 60 mph in 2 seconds?
  8. Oct 30, 2006 #7
    If it were the full force, 455N, it would accelerate the 23.4kg object at 19.5 m/s^2.

    Think about it. The acceleration due to gravity is 9.8m/s^2, correct? That's nearly 22 mph per second. Now, a 229N object in would experience 229N force from gravity, by definition. Soooo, a greater force (hence, a greater than half portion of 455N), would exert a greater acceleration. That's the intuition part that you're arguing with me about. The math works out perfectly as well ;)

    EDIT: That acceleration is without any friction as well. Which is what I believe he was originally aiming at?
    Last edited: Oct 30, 2006
  9. Oct 30, 2006 #8
    Yes, you're absolutely right.

    As for the original poster, I calculated an acceleration of 4m/s/s, not 16. The time is close to my value.
  10. Oct 30, 2006 #9
    The frictional force opposing movement is 0.57*(229 + 261)N
    =279.3 N

    ma = 372.7 - 279.3 = 93.4 newtons,the mass is 229/9.8 = 23.4 Kg

    a=93.4N/23.4Kg = 4 meters/sec^2

    Since at^2/2 = distance ----------> sqrt(2*distance/a) = t = 1.58 sec

    yes yes i finaly got it thanks guys
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook