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!

I can not calculate these newton equations

  1. Jun 29, 2009 #1
    A block B of 20 kg is trapped by a wire of 2 m a car A of 30 kg. Determine (a) the acceleration of the car (b) the tension of the wire, immediately after the system was abandoned the rest, in the position shown in the figure. Despise is the friction.

    Answer:
    (a) 6,17 m/s^2
    (b) 144 N

    My attempt
    CAR

    ZFx = mA . a
    Tcos(65) = 30a

    Block B
    ZFx = mB . a
    Tsin(65) = mB.a

    I can not calculate these equations
     

    Attached Files:

  2. jcsd
  3. Jul 1, 2009 #2
    Re: Newton


    The equations are correct?
     
  4. Jul 1, 2009 #3

    drizzle

    User Avatar
    Gold Member

    Re: Newton

    why are you writing ZFx? do you mean the x component of F? but the car moves in a direction where it has x and y components of the force [Fx and Fy]
    for the tension of the wire there’s no Fx
     
  5. Jul 1, 2009 #4
    Re: Newton

    ZF => Sum of forces
     
  6. Jul 1, 2009 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Newton

    It is probably best to choose the x axis as the axis parallel to the incline, both when looking at the forces on block A and the forces on Block B. If you look at block B first and sum forces in the "x" direction, that is, in the direction parallel to the incline, then it's Tsin25 plus the weight force component =20a, where 'a' is the acceleration in that x direction parallel to the incline. Then look at Block A using the same set of axes, and don't forget to include the weight (gravity) force acting down the incline.
     
  7. Jul 2, 2009 #6
    Re: Newton

    Not Tcos (25) ?
    The acceleration of the block is the same car ?
     
  8. Jul 2, 2009 #7

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Newton

    Sorry, I may have misread this problem. I believe the problem is asking for the car acceleration and the tension in the wire holding the 20 kg mass (Block B) at the instant the car is released from rest, while the wire is in the vertical position. If that is the case, it seems you should look at block B and sum forces in the vertical y direction to get 20g - T = 20a_y. Then look at the car, and sum forces along the incline ( the weight and tension components acting along the incline as the assumed x direction) equal to 30a_x, where a_x and a_y are related by a_x =a_y/sin25.
     
  9. Jul 2, 2009 #8
    Re: Newton

    How to calculate sum of forces in the X axis? Only has the weight
     
  10. Jul 2, 2009 #9

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Newton

    For block B, I took the y axis as vertical and the x axis as horizontal, and summed forces in the y direction. But for the free body diagram of car A, I chose the x axis to be the axis parallel to the incline, that is, 25 degrees up from the horizontal. So acting along the x axis , you have the component of the cars weight (mgsin 25) and also the component of the Tension force (Tsin25). Their sum is 30a_x. You will need to solve 2 equations with 2 unknowns, noting first the relation beteen a_x and a_y (since the block and car are connected, they must accelerate at the same rate).
     
  11. Jul 2, 2009 #10
    Re: Newton

    Thanks
     
  12. Jul 4, 2009 #11

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Newton

    Clarificaton Edit: For block B, I took the y axis as vertical and the x axis as horizontal, and summed forces in the y direction. But for the free body diagram of car A, I chose the x axis to be the axis parallel to the incline, that is, 25 degrees up from the horizontal. So acting along the inclined x axis , you have the component of the cars weight (mgsin 25) and also the component of the Tension force (Tsin25). Their sum is 30a_x. You will need to solve 2 equations with 2 unknowns, noting first the relation beteen a_x and a_y (since the block and car are connected, they must initially vertically accelerate at the same rate a_y, but not so horizontally...until ultimately after a very brief time, once the lower block swings into its final position with the rope perpendicular to the incline, they accelerate together in both the x and y directions, at the same rate, but different from the initial value of the calcualted accelerataion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook