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Homework Help: Hanging mass + acceleration

  1. Mar 25, 2010 #1
    A 50kg weight hangs from the ceiling of a cart that is accelerating horizontally. While the cart is accelerating, the weight swings to the left so that the string makes an angle of 37 degree with the horizontal ceiling. a) What is the size and direction of the acceleration of the cart? b) What is the size of the tension force in the string?

    Fnet = ma
    Fw = mg
    Ftx = Ftcos37
    Fty = Ftsin37

    So first I tried solving for the Fw using Fw = mg. After finding that I assumed that Fw = Ft and used it to solve for Ftx, but then i'm stuck and have no idea what i'm doing.
     
  2. jcsd
  3. Mar 25, 2010 #2
    Such problems are often being observed from the reference frame fixed to the cart. Since the cart is accelerating, such reference frame is non-inertial. 1. and 2. Newton's laws are not valid there, but we can still use them (just formally, in order to have a mathematical description of the problem).

    From such frame you can see that the body you got is hanging from the ceiling forming a certain angle (it is at rest!). Now, only two real forces on the body are gravitational force and string tension. But, according to 1. Newton's net force on the body must be zero for body to be at rest. So we introduce one fictive force (inertial force). Its magnitude is [tex]F_{i}=ma[/tex], where [tex]a[/tex] is acceleration of the cart. This force is always pointing in opposite direction of the cart acceleration. (Remember: this force actually doesn't exist!)

    Now, only thing you need to do is to represent each of the forces you have (gravitational, tension and fictive force) along the x and y axis. Net forces along those axes must be zero.

    So, you have these equations ([tex]T[/tex] is the tension force and [tex]\alpha[/tex] is the angle):

    [tex]T \sin \alpha - ma = 0[/tex] (x axis, right is taken as positive direction)

    [tex]T \cos \alpha - mg = 0[/tex] (y axis, up taken as the positive direction)

    From there, you can easily obtain magnitude of the acceleration and tension force.
     
  4. Mar 25, 2010 #3
    Ok, so I understand that when the cart is accelerating, the hanging mass is at rest and what you propose make sense. So for this case, is the Ft equals to the Fw since the object is at rest? And I don't get this fictive force thing =|.
    Btw, the answers are:
    a) 13 m/s/s to the right
    b) 8.3n

    EDIT: Sorry I read the question wrong, the weight is only .50 kg!!
     
    Last edited: Mar 25, 2010
  5. Mar 25, 2010 #4
    Fictive force is here just to eliminate the effect of other two forces so that the body is at rest. You can also observe the problem without fictive force. Just remember that you are dealing with vectors.

    Then you look at the net forces along each of the axes:

    [tex]ma_{x}=T \sin \alpha[/tex] ([tex]a_{x}[/tex] is acceleration along x axis)

    [tex]ma_{y}=T \cos \alpha - mg[/tex] ([tex]a_{y}=0[/tex] is acceleration along y axis)

    From these two equations you get

    [tex]a_{x}=g \tan \alpha[/tex]

    [tex]T=\frac{mg}{\cos \alpha}[/tex]

    where [tex]\alpha = 90^{\circ}-37^{\circ}=53^{\circ}[/tex] (I messed up the angle a bit because I didn't read the text of the problem well, sorry.) [tex]a_{x}[/tex] is the acceleration you are looking for (the acceleration of the body looking from the inertial frame of the Earth, not from the cart). Acceleration of the body is equal to the acceleration of the cart (again, if you look from outside the cart).

    Sorry if I confused you with too much theory, but I think it is important to know what's going on.
     
  6. Mar 25, 2010 #5
    Thank you! I understand now.
     
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