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: Force (object is suspended connected to a block on table and pulled up)

  1. Oct 20, 2007 #1

    ~christina~

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    A block of mass m1 on a rough horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight frictionless pulley as shown. A force of magnitude F at an angle theta with the horizontal is applied to the block. The coefficient of friction between the block and surface is [tex]\mu_k[/tex]

    a.) draw a free body diagram of the 2 masses

    b.) use newton's dynamic equation to determine the magnitude of the acceleration of theh 2 objects in terms of their masses, F, theta, and [tex]\mu_k[/tex]

    c.) what is the tension in the cord
    http://img518.imageshack.us/img518/1023/picture1dn4.png [Broken]
    2. Relevant equations

    F= ma

    Ffk= uk* Fn

    and a newton's dynamic equation? ==> what is that exactly??

    3. The attempt at a solution

    a)for the free body diagram

    http://img484.imageshack.us/img484/2900/90134448jw8.th.jpg [Broken]

    b) how would I do this question...not sure and I'm stuck after the free body diagram...hm..


    c) I guess I would need to solve part b to get part c


    ~I do know that they both have the same accelration but how would I know if the total forces [tex]\sum Fx= 0[/tex]?

    I need help please..
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 20, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's Newton's 2nd law.

    One problem with the FBD: That applied force F is at an angle, not horizontal.

    Analyze the forces parallel to the direction of motion for each block and apply Newton's 2nd law. You'll get two equations--one for each mass--that you'll solve together for the two unknowns: the acceleration and tension.
     
  4. Oct 21, 2007 #3

    ~christina~

    User Avatar
    Gold Member

    I was wondering why I don't analyze the horizontal component of the block but I think it's b/c the total forces = 0 but...this is what I got.

    for m2 (only movement in y direction)
    [tex]\sum Fx= 0[/tex]
    [tex]\sum Fy= T- m_2 g= m_2 ay[/tex]

    for m1 (block)
    [tex]\sum Fx= max= Fcos theta- (T+ \muk m1g)[/tex]
    [tex]\sum Fy= may= Fsin theta + N- mg= 0[/tex]

    Well I just wanted to check first...
     
  5. Oct 21, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Good. I would just call the acceleration "a".

    Careful here: What's the normal force? Again: call the acceleration "a".
    Good. Use this to figure out N.
     
  6. Oct 21, 2007 #5

    ~christina~

    User Avatar
    Gold Member

    for m2 (only movement in y direction)
    [tex]\sum Fx= 0[/tex]
    [tex]\sum Fy= T- m_2 g= m_2 a[/tex]

    for m1 (block)
    [tex]\sum Fx= ma= Fcos theta- (T+ \muk m1g)[/tex] => was this incorrect (see below for my correction)??

    ~I do know that that is the Fx component right? but I guess I need to incorperate the N?


    [tex]\sum Fy= ma= Fsin theta + N- mg= 0[/tex]
    from the Fy

    N= mg - Fsin theta


    I think that since the Ffk = uk*FN
    then the equation on the top for the Fx wouldn't be uk* m1g since the normal force is lighter so it would be instead

    [tex]\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))[/tex]


    Is it fine now?

    I guess whenever I have to find the x component of the force when there is friction I need to incorperate the normal force the angle it is being pulled at if there is a angle so that the normal force is not just m*g right?
     
  7. Oct 21, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, all better.

    Right. Friction depends on the normal force, which may or may not equal the weight of the object, depending upon the situation.
     
  8. Oct 21, 2007 #7

    ~christina~

    User Avatar
    Gold Member

    [tex]\sum Fy= T- m_2 g= m_2 a[/tex]

    [tex]\sum Fy= ma= Fsin theta + N- mg= 0[/tex]
    from the Fy

    N= mg - Fsin theta

    [tex]\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))[/tex]

    ________________________________________________________________
    but how would I find the part b) which is...

    b) b.) use newton's dynamic equation to determine the magnitude of the acceleration of the 2 objects in terms of their masses, F, theta, and [tex]\mu_k[/tex]

    (I'm not sure what they want)

    would it be adding the 2 equation??

    the first object and second object's forces like this?

    [tex]\sum Fy= T- m_2 g= m_2 a[/tex]

    [tex]\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))[/tex]

    so adding them I get..
    + T- mg= m2a
    + Fcos theta - T- uk (m1g- Fsin theta) = m1a
    ______________________________________

    Fcos theta- m2g- uk(m2g- Fsin theta)= (m1 + m2)a

    a= (Fcos theta- m2g - uk(m2g- Fsin theta))/ m1+ m2
    Is this what they want?


    Assuming it is...
    for part c.) I need to find

    c.) tension in cord..

    I think that I would plug this into a original equation while using the acceleration I found...

    [tex]\sum Fy= T- m_2 g= m_2 a[/tex]

    T= m2a + m2g

    substituting a

    T= m2( (Fcos theta- m2g - uk(m2g- Fsin theta))/ m1+ m2) + m2g


    I think this is okay but not sure..
     
  9. Oct 21, 2007 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me!

    As a practical matter, I would solve for a and then plug the numerical value into the other equation to get T.
     
  10. Oct 21, 2007 #9

    ~christina~

    User Avatar
    Gold Member

    Oh okay..I would plug in the numbers if I had numbers given...

    I usually find it challenging to not have numbers though..

    Thanks for your help Doc Al !
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook