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: Online Course Work Question - Dynamics

  1. Sep 25, 2006 #1
    Hello everyone, I'm sorry if a question like this has been posted before, but I am trying to get my online work done for tonight and I have come across two problems that I just cant get past. I have gotten far into the problems but the website I enter values in on says theyre wrong. I appreciate any help I can get.

    Question 1:


    On a level road with its brakes on, the shortest distance in which a car traveling with 98 km/hr can stop is 92 m. The shortest stopping distance occurs when the driver uses anti-lock brakes, which means that the car brakes without skidding.

    I found |a| to be: 4.027 m/s^2

    Next it asks for b) What is the coefficient of static friction between the tires and the pavement?

    I looked up the coefficient of friction for concrete and rubber and it says 1. Now when I put that in it says it's wrong. I was wondering how its possible to figure out a problem like this with no value for mass.

    Question 2:


    A block of mass M = 15 kg is suspended at rest by two strings attached to walls, as shown in the figure. The left string is horizontal and the right string makes an angle q = 50° with the horizontal. What is the tension in the left string? (Assume the ring where the strings come together is massless.)

    Ok, this one I thought I had the right answer after working through it but the site says I'm wrong. Here's what I did:


    Since there are two strings holding up the block, 147.15/2=73.58

    That means that the T in the y direction of the right string is 73.58N.

    Then TRy=73.58/sin50 or 96.045 N

    Now if I might be mistaken but that force doesnt really matter because what I can use is the T on the right string in the x direction is equal to the T on the left string in the x direction (because the system is in equilibrium).

    So: F in x direction on the right string= 73.58/tan50

    Which means that the T on the left string= 61.74 N

    I might be missing a key concept here, but any help with this would be appreciated. Thanks! (And go easy on my please, I know what its like to read posts by newbies.)

    Attached Files:

  2. jcsd
  3. Sep 25, 2006 #2
    Yeah, Ive been looking around online and I havent seen any equation or group of equations to figure out friction without a mass.
  4. Sep 25, 2006 #3

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    You have the acceleration. Newton's 2nd law says what about acceleration? (hint: consider the frictional force to be the only force on the car tires)
  5. Sep 25, 2006 #4

    Ok, I was trying to use that but dont we have two unknowns here?

    We dont know m, and we cant get the FF=(mu)FN

    So how can I use the Force of friction on the tires if I have no mass and only acceleration?
  6. Sep 25, 2006 #5
    Oh, and there's an update...I got the second problem by not dividing the force in the y direction by two. It was a pure guess and check, so if someone can explain why that would be greatly appreciated.
  7. Sep 25, 2006 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    Chi meson was alluding to m*a = (mu) m*g => a = (mu) g and knowing a and g, one calculates 'mu'.

    For the second problem, one string holds up the block. The vertical component of the Tension in the string at 50° supports the weight of the block. The horizontal component of the Tension in the string at 50° is balance by the Tension in the horizontal string.
  8. Sep 25, 2006 #7
    Oh wow, I cant believe I didnt see that, I had it almost written like that except I dint have ma on the other side, in which case i would have no way to cancel the mass.

    Thanks much both Chi and Astro!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook