# Finding spring balance readings (newtons)

1. Oct 2, 2007

### dnt

1. The problem statement, all variables and given/known data

Find the readings on each of the balances in these 4 situations (strings are massless and incline is frictionless):

(sorry about the photo - dont know how to fix it)

2. Relevant equations

F=ma, W=mg

3. The attempt at a solution

1) Seems easy enough. Im pretty sure the balance would just read 98 N since the tension above equals the weight of the mass below.

2) Im having a little trouble with this one. I cannot tell if its going to be 196 N (98 N from each of the weight pulling down on each end) or just 98 N. Each of the blocks has a net force equation of:

F = W - T = ma

98 - T = 0

T = 98 N (T, the tension, is what the scale will read, correct? Thats where the answer comes from?)

so I get the tention in the string is 98 for both but I dont know how to combine the two net force equations into a single answer for the balance reading.

3) the block has a net force equation of:

F = W - T - T = ma

W - 2T = 0

W = 2T

so the tension in the string/balance reading would be half of the blocks weight = 49 N.

4) I know I need the component of the blocks weight on this one due to the incline. And that should be Wsin30 = 49.

so the net force equation here is:

F = Wsin30 - T = ma

T = Wsin30 = 49 N

is that the correct answer? What about that other 10 kg block above? How does it get included into the problem (if at all?)

thanks.

2. Oct 2, 2007

### Staff: Mentor

Looks like you got them all correct!

Regarding (b), realize that the force pulling each side of the spring balance is always the same. It's really no different than (a).

Regarding (d), that upper 10 kg mass is irrelevant, since it's tied to some support. Again, the amount of force pulling on each end of the spring balance is the same: 49 N.

3. Oct 2, 2007

### learningphysics

If you're unsure, this is the way I would look at it... examine the freebody diagram of the little hook thing. On one end there is the tension in the rope acts on the hook. On the other end the force of the spring acts on the hook. The two must be equal and opposite because the hook isn't moving (acceleration is 0... also it is massless... either one alone is enough for Fnet = 0).

So the tension in the rope attached to the hook is equal in magnitude and opposite in direction to the force the spring exerts... and hence the weight that is read by the scale equals the tension in the rope.

4. Oct 2, 2007

### dnt

thanks for the help guys.

just so i fully understand part (b), how would you show the work for it? im still having trouble writing a net force equation using both blocks (all i did was write it for one block). it seems different than (a) even though its not. ie, why is it not greater since both blocks are pulling on it?

im having trouble getting part (b) into my head...this one is confusing and i want to make sure i understand it 100%. thanks again.

5. Oct 2, 2007

### learningphysics

A spring can be seen as a symmetrical object. Hooke's law says that a spring exerts a force of F = kx. But at which end is the force exerted? The answer is both ends... If a spring is horizontal and stretched by an amount x from the unstretched state... then one end exerts a leftward force of kx... and the other end exerts a rightward force of kx. This is necessary for the system to be in equilibrium. (if the spring exerts a leftward force of kx... that means it has a rightward force of kx acting on it by newton's 3rd law... if the spring exerts a rightward force of kx, then by newton's 3rd law something exerts a leftward force of kx on it... so the two forces on the springs balance).

So you just need to examine one end. It doesn't matter which one (as long as the system is in equilibrium. in all these situations we have equilibrium).

For the first diagram... take the freebody diagram of the little hook. What are the forces acting on it?

6. Oct 4, 2007

### dnt

just gravity i believe. im starting to get it i think. it just seems that (with the 2nd one) each weight is pulling on the spring so the reading should be double - 100 N pulling on one end, and another 100 N pulling on the other.

7. Oct 4, 2007

### learningphysics

gravity downwards... and the spring force upwards...

the second one isn't different from the first. remember in the first the ceiling is also exerting an upwards force of 10*g... so even in the first one both ends ends are being pulled at...