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: Help! Problem on INCLINED planes.

  1. Mar 25, 2006 #1
    Hello, here is the question.

    Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope over a “frictionless” pulley as pictured in the diagram. The ramp is inclined at 30.0º and the coefficient of kinetic friction on the ramp is 0.18.


    (b) Determine the acceleration of the system once it begins to slide.

    i only need help on b.. finding the acceleration. if i can get the right acceleration i can get all the other parts.

    Now i solved it by doing this. making 2 equations.

    4 kg = m1 6kg = m2 motion to the right is positive. uk = 0.18

    m1a = T - [Ff + Fg||]
    m1a = T - (uk)(m1g)cos30 - sin30m1g
    m1a = T - 6.1 - 19.6 equation 1

    m2a = m2g - T
    m2a = 58.8 - T equation 2

    add the two equations together you get. (T cancel out)

    a (m1 + m2) = 58.8- 25.7
    a = 33.1 / 10
    a = 3.3 m/s ^2

    i have been lookin over my calculation over and over and the acceleration is different from the answers.

    here is the SOLUTION:

    For the 4.0-kg mass:

    4.0 kg(a) = T – uKmg(cosX) – mg(sin X)

    4.0 kg(a) = T – 13.5 N

    For the 6.0-kg mass:


    6.0 kg(a) = 58.8 N – T

    Solving the system of equations:

    a = 4.5 m/s2

    according to the solution... for equation 1.. it is + 6.1 - 19.6 = -13.5

    but it doesnt make sence because it should be -6.1 - 19.6 = -25.7

    is there a trick im missing or do you think the solution is wrong? this question is vital for a test on monday.

    thank you

  2. jcsd
  3. Mar 25, 2006 #2


    User Avatar
    Homework Helper

    You are correct, friction always opposes motion. The signs used in the accepted solution are wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook