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: Tension problem on an inclined plane

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Block 1 of mass m1=2.0kg and block 2 of mass m2=3.0kg are connected by a string of negligible mass and are initially held in place. Block 2 is on a frictionless surface tilted at theta = 30 degrees. The coefficient of kinetic friction between block 1 and the horizontal surface is .25 The pulley has negligible mass and friction. Once they are released, the blocks move. What then is the tension in the string?

    Please see the attached diagram I drew in paint.

    2. Relevant equations

    F = ma (newtons second law)

    3. The attempt at a solution

    I worked out a solution but I wanted someone to double check my answer because I've been having problems with this:

    T = Tension
    fk = kinetic friction

    Equation for Mass 1: T - (m1)(g)(fk) = (m1)(a)
    Equation for Mass 2: (m2)(g)(cos30) - T = (m2)(a)

    Combined: (m1)(a)+(m1)(g)(fk) = (m2)(a)+(m2)(g)(cos30)

    Plugging in the masses, g = 9.8 and fk = .25 I come up with a = 4.112

    Plugging a back into the equation for mass 1 gives me T = 13.124 N

    Thanks for any help. I feel like I'm missing part of it or I need to initially set a to 0?

    Attached Files:

  2. jcsd
  3. Mar 10, 2010 #2


    User Avatar
    Homework Helper

    Ηι chronos12, welcome to PF.
    Equation for Mass 2: (m2)(g)(cos30) - T = (m2)(a)
    Check this equation. In the diagram you have not shown θ.
  4. Mar 10, 2010 #3
    Please see the attached revised diagram showing the angle. There should be a negative sign in front of (m2)(a). On paper I wrote this down - just didn't type it correctly here. The combined equations from my work up should read: (m1)(a)+(m1)(g)(fk) = -(m2)(a)+(m2)(g)(cos30).

    When you say check equation 2, I'm thinking this should have been sin(30) instead of cos(30)?

    It would then read: (m2)(g)(sin30) - T = (m2)(a)

    Combined: (m1)(a)+(m1)(g)(fk) = -(m2)(a)+(m2)(g)(sin30)
    The new value for a = 1.96
    The new value for T = 8.82 N

    Please let me know if I'm correct on this.

    Sorry for the bad paint skills, and thanks for the reply!

    Attached Files:

    Last edited: Mar 10, 2010
  5. Mar 10, 2010 #4


    User Avatar
    Homework Helper

    Component of g along the inclined plane is g*sinθ, not g*cos θ
  6. Mar 10, 2010 #5
    Thanks for the revision!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook