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Spring scale- confusing me help

  1. Aug 26, 2006 #1

    firstly Thanks for the help.....
    I am not very good at forces ... and spring balances etc....

    Q1 A 10N force is pulling up on the ring of spring scale that weighs 2N. If an 8N mass is attached to the bottom hook of the scale, what is the scale reading?

    I figure that the mass of the spring is not relevent - hence the answer should be 10+8 = 18N. But the answer is 8N !

    Hence I can't solve the next one also:
    Similar as Q1 above but this time ithe force pulling up the ring of the spring scale is 5N instead....

    I am really confuse about these. please help....

  2. jcsd
  3. Aug 26, 2006 #2

    Doc Al

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    What the spring scale reads is the tension exerted by its bottom hook (where you hang things that you want to weigh). The first thing to do is to examine all the force on this system (scale + mass) to find out what it's doing. There's an upward force of 10N, and the downward forces of the weight of the scale (2N) plus the weight of the mass (8N). Those add up to zero, so you know the acceleration is zero. So... what force must the hook be exerting to support the hanging mass?

    This time the forces do not add to zero. Hint: Find the acceleration of the system. Then apply Newton's 2nd law to the hanging mass to find out what force the scale exerts on it.
  4. Aug 26, 2006 #3
    The first bit is simpler than it sounds, matthew, maybe it's rather a trick question. Imagine you're holding up the scale, using a force of 2N because that's what it "weighs". Now you hang another 8N on the hook, and you're now exerting 8+2=10N. But the scale says 8N because that's whats on the hook.

    Actually, the second bit is easier than it sounds too. I won't tell you the answer. But here's a clue: equal and opposite.
  5. Sep 2, 2006 #4
    Ok, let me try

    Q1 - there is an upward force of 10N. there are two downward forces - the spring 2N and the stone 8N. So Nett force is zero. Hence the spring reading is the reading of the stone which is 8N. am I right in my understanding?

    Q2. Upward force = 8N,
    Downward force = 2N (mass spring) + 8N (stone)
    Nett force 10-8N = 2 N downward
    so the spring reading is 2N

    Somehow, it doesn't feel right.
    Please help me to correct....

    cheers and thanks
  6. Sep 3, 2006 #5

    Doc Al

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    Sounds good to me.

    The upward force is given as 5N, not 8N. The downward force--the weight of scale plus hanging mass--is the same. So what's the net force on the system? What's the acceleration of the system?

    Use that to deduce the force that the scale must exert on the hanging mass. Which is also the force that the hanging mass exerts on the scale--via Newton's 3rd law. And that's the force that the scale will read.
  7. Sep 3, 2006 #6
    Try this again:

    Q2. Upward force = 5N, Mass spring = 2N, Mass stone = 8N

    Downward force = 2N (mass spring) + 8N (stone) = 10N
    Nett force 10-5N = 5 N downward
    so the spring reading is 5N

    Is this correct?

    Thanks a lot.
  8. Sep 4, 2006 #7

    Doc Al

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    Do exactly as I said. First find the acceleration of the system using Newton's 2nd law.

    Hint: You know the weight of the scale and the hanging mass--but what are their masses?
  9. Sep 4, 2006 #8

    weight of scale = 2N => mass = 0.2kg
    weight of stone = 8N => mass = 0.8 kg
    total mass = 1.0kg

    now what....

    a = F/m

    which force do I use to find a?

    5 N upwards pulling force ? a = 5 m/s2 upwards

    or use nett force 5N down a= 5 m/s2 downswards

    either way .... what do I do next?

    thanks for defusing my confusion......
  10. Sep 4, 2006 #9

    Doc Al

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    To find the acceleration of the scale + stone, use the net force acting on it. Once you have the acceleration, analyze the forces acting on the stone. (Hint: There are two forces acting on the stone.)
  11. Sep 4, 2006 #10

    using the nett force 5N downwards , a = 5 m/s2 down.
    the stone has an upward force of 5N and its own weight 8N down.
    Nett force on stone is F = 8-5 = 3N down.

    So the mass reaading on scale is F/a = 3/5 kg??

    Am I getting anywhere ???
  12. Sep 4, 2006 #11

    Doc Al

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    No. The 5N force is the upward force on the scale, not on the stone. The upward force on the stone is what you are trying to figure out. But, yes, the downward force on the stone is its weight, 8N.

    Since you know the stone's mass and acceleration, what must be the net force on it?
  13. Sep 7, 2006 #12
    Mass of stone = 0.8kg,
    Force Nett = 5 m/s2 down.
    So the reading on the scale = F = ma = 0.8 x 5 = 4 N.

    Am I right?

  14. Sep 8, 2006 #13

    Doc Al

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    The acceleration of the stone is 5 m/s^2 down. Thus the net force on the stone is F_{net} = ma = 0.8 x 5 = 4 N.

    The scale doesn't read the net force, it reads the force that it exerts on the stone. Let's figure it out.

    What forces act on the stone? There are two:

    (1) The scale force acting upward, which is what we are trying to find. Let's call that F_s.

    (2) The weight acting downward, call it F_w, which equals 8 N.

    Calling up to be positive, what's the net force in terms of these two forces? F_{net} = F_s - F_w

    But we already calculated F_{net} above to be -4 N.

    You do the last step: plug in the values and solve for F_s.
  15. Sep 9, 2006 #14
    Thanks. Trying.....

    Force on hook of scale = 5N up
    Weight of scale = 2 N down ===> mass of scale = 0.2kg
    Weight of stone = 8N down ====> mass of stone = 0.8kg
    Force nett of above = 10N-5N = 5 N down.

    Mass of scale and stone = 1 kg
    Hence Nett acceleration of system = F/m = 5 m/s2 down.
    Hence Force nett on stone F(nett) = ma = 0.8kg x 5 m/s2 = 4N down.

    If up is +ve, then

    F(net) = -4 N
    F (on weight if stone) = - 8N (given)

    Forces On stone:

    F(nett) = F (scale pulling stone up) - F (force on stone mass)
    -4 = F (scale) - (-8)
    F (scale) = -4 - 8 = -12 N.
    Hence reading from scale = -12N

    How am I doing now ????
  16. Sep 9, 2006 #15

    Doc Al

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    All good.


    F(net) = F (scale; up) + F (weight; down)
    -4 = F (scale) + (-8) = F (scale) - 8

    Almost there. :wink:
  17. Sep 9, 2006 #16
    F(scale) = 8-4 =4N

    But earlier it was
    F(net) = F(scale; up) - F(weight; down)

    why "-" is changed to "+" now?


    PS: can u post a similar Q so that I can test if I really understand this....

  18. Sep 9, 2006 #17

    Doc Al

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    Nothing has changed. The net force is always the sum of all forces acting on an object. Forces that act upwards are positive; those that act downward (like weight) are negative. (Assuming a sign convention that uses positive to indicate "up", of course.)
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