# I don't understand this dynamics question?

1. Apr 19, 2010

### Cokebottle

A spring balance reads 20N when a stationary object is suspended. Accelerating at 5m s^-2, What is the reading as the object when it is

(i) ascending
(ii) descending

I don't understand this question at all..

Can someone explain to me how do I solve it?

2. Apr 19, 2010

### Filip Larsen

Welcome to PF.

A spring balance uses a spring to counter the weight of an object (lets call it a mass), so when the mass is at rest the force generated by the spring due to it being deflected (that is, compressed if the mass is placed "above" the spring or extended if the mass is "hanging" from the spring) precisely balances out the force of gravity on the mass. The weight of the mass can then be read of a scale according to deflection of the spring.

Using Newtons 2nd law, this situation can be described by saying that the sum of the spring and gravity force is zero, i.e. Fs+Fg = ma = 0 because a is zero.

If you then accelerate the whole spring-mass up or down and still require the setup to be in balance, the two forces no longer sum to zero since the acceleration in the above equation now is plus or minus the given 5 m/s2. So, to figure out how much deflection the spring now must have you can insert what you know about the forces Fg (based on mass m and constant g) and Fs (based on spring constant k and deflection x), and you can solve to get the deflection.

3. Apr 19, 2010

### dddaaakkk

Hello Cokebottle,

I think you should consider process of oscillation of the object on the spring. The dynamic equation will be
ma=mg-T
where m is the mass of the object, a is the acceleration of the object (is positive when is downward-directed and negative otherwise), g is the free fall acceleration and T is the spring force and it is T that responsible for value of reading of the spring balance.
So when the system is calm a=0 and mg=Tcalm=20 N
When the object is ascending we can infer that the acceleration is downward-directed (because the acceleration is always contra-directed to the velocity in oscillations), so
T=m(g-a)=mg(1-a/g)=Tcalm(1-a/g)=10N (approximately)
When the object is descending we can infer that the acceleration is upward-directed, so
T=m(g+a)=mg(1+a/g)=Tcalm(1+a/g)=30N (approximately)

Last edited: Apr 19, 2010