1. Not finding help here? Sign up for a free 30min 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!

Tension with a tension problem?

  1. Oct 29, 2004 #1
    "tension" with a tension problem?

    11. [CJ6 4.P.076.] A 248 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.880, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.

    I dont know where to start please help me out!!!!!!!!!!!!!!!!!!!!!! :

    surprised :cry: :surprised :cry:
  2. jcsd
  3. Oct 29, 2004 #2
    First of all, let us discuss the concept of the tension. Essentially, the tension is the force with which the block is being pulled. Therefore, the forces accross the "x" plane are the tension and the friction forces, as well as the "x" component of the gravitational force. On the "y" plane, we find the normal force as well as the "y" component of the gravitational force.
    (note that I rotated our x-y plane by 29 degrees, so that all the forces are at 0 degrees to the surface )

    Now, lets write what we've just said down:
    [tex]F_{net,x} = F_t - F_k - F_{g,x}[/tex]
    [tex]F_{net,y} = F_n - F_{g,y}[/tex]
    To find [tex]F_k[/tex], we want to find the normal force. Since we know that no acceleration on the "y" plane exists (that is, the log doesn't levitate above the surface), [tex]F_{net,y} = ma = m(0) = 0[/tex]. So, [tex]F_n = F_{g,y}[/tex]. Looking at the block diagram you should've drawn, you should get [tex]F_{g,y}=F_n=mg sin{69^0}[/tex].
    Let's go back to friction. [tex]F_k = \mu _k F_n = \mu _k (mg sin{69^0} )[/tex].
    We substitute into our original [tex]F_{net,x}[/tex]:
    [tex]m a_x = F_t - \mu _k (mg sin{69^0} ) - (mg cos{69^0})[/tex]
    [tex] F_t = m a_x + \mu _k (mg sin{69^0} ) + (mg cos{69^0})[/tex]
    Substitute, and you're done.

    Hope this helps,
  4. Oct 29, 2004 #3
    still tryin

    Thanks for the help! I followed your steps above however the answer i got is not the correct answer. Am i doing the problem incorrect.

    F=2367.23 is not correct
    could you please tell me what i am doing wrong. :confused:
  5. Oct 30, 2004 #4
    Where did 69 come from? The ramp is on a 29 degree incline. Charvonne, i don't think you quite understand the problem, because although evgeny did make a mistake, his process is right. That is something you should be able to pick up on and figure out the solution.
  6. Oct 30, 2004 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Actually, Evgeny made two mistakes. One, he used 69 degrees instead of 29. But he also mixed up sine and cosine. For example, the normal force is [itex]mg cos\theta[/itex], not [itex]mg sin\theta[/itex]. Similarly, the component of the weight down the ramp is [itex]mg sin\theta[/itex], not [itex]mg cos\theta[/itex]. (Where [itex]\theta[/itex] = 29 degrees.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Tension with a tension problem?
  1. Tension problems (Replies: 8)

  2. Tension Problem (Replies: 2)

  3. Problem on tension (Replies: 4)