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Lamp Hanging in an Elevator

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.6 m/s2. (a) If the tension in the cord is 92 N, what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of 1.6 m/s2?


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    a) using M for mass, so as not to confuse it with 'm' for meters
    F(net)=M*a, T-F(gravity) = M*a
    92N - M*(9.8 * m/s^2) = M*(-1.6 * m/s^2)
    9.8*M - 1.6 M = 92 kg
    M = 11 kg

    b)
    T - (11 kg)*(9.8 m/s^2) = (11 kg)*(1.6 m/s^2)
    T = 18N + 110N
    T = 128N

    Both of my answers are wrong. Where am I going wrong in my thought process?
     
  2. jcsd
  3. Feb 15, 2012 #2
    Always look at your equation from a common sense perspective. You have an elevator that is going down and slowing its descent. That tells me that if you are looking at tension in the cord that suspends it and the acceleration terms are on the same side of the equation, they had better have the same sign so that they will add together in a positive manner to increase tension.
     
  4. Feb 15, 2012 #3
    In part a) the elevator is decelerating downwards, so I made it negative. In part b) the elevator is accelerating upwards, so I made it positive. My thought process was:

    Upwards (positive) vectors:
    Tension T

    Downwards (negative) vectors:
    M*Gravity

    acceleration = 1.6 (part b only)
    deceleration = -1.6 (part a only)

    I'm sorry, but I'm not entirely sure what you're trying to tell me.
     
  5. Feb 15, 2012 #4
    92N - M*(9.8 * m/s^2) = M*(-1.6 * m/s^2)

    Above is your equation. If you move the second term on the LHS to the RHS you get

    92N = M*(9.8 * m/s^2) + M*(-1.6 * m/s^2)

    As you can see, they don't have the same sign. Since the elevator is slowing and going down, the tension (which is due to weight and acceleration) must get greater. How can it with your signs being different?
     
  6. Feb 15, 2012 #5
    So I don't need to negate the acceleration even though it's decelerating, because it's not an actual force vector acting upon the lamp? I removed the (-) and my answer is correct now (thank you!), but I just want to understand the 'why' so I can not make this mistake again.
     
    Last edited: Feb 15, 2012
  7. Feb 15, 2012 #6
    Draw a free body diagram of the lamp (descening elevator) calling it M for its mass. There are several forces on it. Use arrows to denote their direction.

    First is its weight; the arrow should be down.
    Section is the tension of cord; arrow should be up.
    Third is the acceleration force due to its slowing down. Because the acceleration is upward (slowing down), the force is in the opposite direction which is down.

    So summing forces and setting to zero with up being positive you have

    T - W - M*a = 0

    where W and M*a are > 0.

    Always look at your equation and see if it makes good sense to you.
     
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