What is the formula for calculating force using a spring balance?

In summary, the student attempted to solve a problem involving a spring balance, but was unable to come up with a solution. In order to solve the problem, they first had to think about the forces acting on each object and draw the free body diagrams. They also had to think about what would happen if a spring was hung from the ceiling and a mass was hung from the spring. They then had to consider the tension in the strings and write an equation to describe the motion of the system.
  • #1
Suyash Singh
168
1

Homework Statement


upload_2018-4-14_9-57-39.png


Homework Equations


only basic school level physics will be needed in this question.

The Attempt at a Solution


The mass m1 pulls spring from first side and m2 pulls spring from the other side therefore reading should be m1-m2.
But my answer is wrong.
I can't understand what to do.
 

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  • #2
Start with the forces acting on each object. Draw the free body diagrams. Also think about what happens if a spring is hung from the ceiling and a mass m is hanging from the spring. what are all the forces on the spring?
 
  • #3
As Chandra posted, draw a free body diagram for each suspended mass.
Think about the tensions in the strings. Are they the same? If not, why not?
Is the system static?
 
  • #4
upload_2018-4-14_20-29-13.png
tension in both strings connected is (m1-m2)g. :(
 

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  • #5
Suyash Singh said:
View attachment 224049 tension in both strings connected is (m1-m2)g. :(
for second body diagram(m2) we replace all 1 with 2 and vice versa
 
  • #6
Why are you stating that the upward force on m1 is m2g?
 
  • #7
Chandra Prayaga said:
Why are you stating that the upward force on m1 is m2g?
because the other mass is pulling it up
 
  • #8
So, is there no difference between this problem and another, in which there is no spring at all, but a single string going from m1 to m2 over the two pulleys?
 
  • #9
Chandra Prayaga said:
So, is there no difference between this problem and another, in which there is no spring at all, but a single string going from m1 to m2 over the two pulleys?
ummm i don't know.We never have practicals.
supposing the spring balance canceled out the m1g force.
tension = m1g ( for string connected to m1)
 
  • #10
Well, it is a simpler problem than the one you started. If you like, you can draw the diagram for that problem, and you will see the point that I am trying to make. That is, we cannot jump to conclusions from every formula that we have on hand. In your situation, you drew the force diagram for only one object. What about the force diagram for the spring?
 
  • #11
You forgot to answer this question:

haruspex said:
Is the system static?
 
  • #12
Suyash Singh said:
tension in both strings connected is (m1-m2)g. :(
You do not know that yet. Just assign unknowns to the two string tensions, and do not assume the system is static.
Post at least one standard physics equation which you believe may be relevant.
Consider the three bodies separately, and apply that equation to each.
 
  • #13
upload_2018-4-15_9-29-43.png
 

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  • #14
haruspex said:
You do not know that yet. Just assign unknowns to the two string tensions, and do not assume the system is static.
Post at least one standard physics equation which you believe may be relevant.
Consider the three bodies separately, and apply that equation to each.
T2=m2g
T1=m1g
spring balance strings have no tension because they are perpendicular
 
  • #15
Suyash Singh said:
T2=m2g
T1=m1g
No. For the third time, do not assume the system is static.
 
  • #16
Suyash Singh said:
spring balance strings have no tension because they are perpendicular
Strings have this clever trick of carrying tension around corners and over pulleys.
 
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  • #17
haruspex said:
No. For the third time, do not assume the system is static.
t1<m1g
t2<m2g
 
  • #18
haruspex said:
No. For the third time, do not assume the system is static.
oh i just saw photo of spring balance
is it like that the weight of only one mass is being measured and the other mass is to hold the spring balance?
 
  • #19
Suyash Singh said:
oh i just saw photo of spring balance
is it like that the weight of only one mass is being measured and the other mass is to hold the spring balance?
if its so then,
acceleration of spring balance=a=(m1-m2)g/m2
 
  • #20
oh and if both masses were equal

both mass=m
then reading of spring balance would be m
 
  • #21
Suyash Singh said:
is it like that the weight of only one mass is being measured and the other mass is to hold the spring balance?
No.
Suyash Singh said:
acceleration of spring balance=a=(m1-m2)g/m2
The spring balance we are told is massless, and we can assume that its reading is constant. So its only functions in the question are to connect the two strings and to measure the tension in them. So apart from the measurement you can ignore the balance and treat the two strings as one.
Let the tension in that one string be T. Write two motion equations, one for each mass.
 
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  • #22
haruspex said:
No.

The spring balance we are told is massless, and we can assume that its reading is constant. So its only functions in the question are to connect the two strings and to measure the tension in them. So apart from the measurement you can ignore the balance and treat the two strings as one.
Let the tension in that one string be T. Write two motion equations, one for each mass.
ok let acceleration of system be a(downwards of m1)

T=M2(g+a)+m1(g-a)
T=m2g+m2a+m1g-m1a
=g(m2+m1)+a(m2-m1)
 
  • #23
Suyash Singh said:
T=M2(g+a)+m1(g-a)
If two teams pull on a rope in a tug of war, each exerting 1000N of force, what is the tension in the rope?
 
  • #24
Suyash Singh said:
T=M2(g+a)+m1(g-a)
What standard equation are you using to get that, and to what free body are you applying it?
 
  • #25
jbriggs444 said:
If two teams pull on a rope in a tug of war, each exerting 1000N of force, what is the tension in the rope?
1000N because they won't cancel out.
 
  • #26
haruspex said:
What standard equation are you using to get that, and to what free body are you applying it?
i am applying it to both the masses

g is downward acceleration due to Earth
a is the acceleration produced in the system due to m1>m2.
a for m2 is in the upward direction
a for m1 is in the downward direction
 
  • #27
Suyash Singh said:
1000N because they won't cancel out.
Right. So why do you think that the forces add in the situation for this problem?
 
  • #28
Suyash Singh said:
i am applying it to both the masses
I asked what standard equation you are using. Please state it.
Apply it to one free body at a time. I don't understand how you are applying it to both at once, but the equation you get is wrong.
 
  • #29
haruspex said:
I asked what standard equation you are using. Please state it.
Apply it to one free body at a time. I don't understand how you are applying it to both at once, but the equation you get is wrong.
standard equation:
Force= mass x acceleration
Tension = force
Applying on m2
force=m2(g+a)
Applying on m1
force=m1(g-a)

where net acceleration is in the downward direction of m1 since m1>m2

now i am confused with tension.
 
  • #30
jbriggs444 said:
Right. So why do you think that the forces add in the situation for this problem?
i don't know the math here but in this case one of the force acts as a "support" ,that is, we could replace one of the force with a wall.

what i mean to say is that two teams pulling rope from two sides with same force
is the same case as
a team pulling a rope attached to a stationary wall with the same force.
 
  • #31
Suyash Singh said:
i don't know the math here but in this case one of the force acts as a "support" ,that is, we could replace one of the force with a wall.

what i mean to say is that two teams pulling rope from two sides with same force
is the same case as
a team pulling a rope attached to a stationary wall with the same force.
but i don't know the math because its not given properly anywhere.
 
  • #32
Suyash Singh said:
Applying on m2
force=m2(g+a)
Applying on m1
force=m1(g-a)
Right. You have two equations and two unknowns. Solve to find the force.
How does that force relate to what the spring balance shows?
 
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  • #33
haruspex said:
Right. You have two equations and two unknowns. Solve to find the force.
How does that force relate to what the spring balance shows?
a=(m1-m2)g/m1+m2
force=2m1m2/m1+m2
oh so the spring balance shows 2m1m2/m1+m2 since it divides the tension produced by g to get the mass
 
  • #34
and
Suyash Singh said:
a=(m1-m2)g/m1+m2
force=2m1m2/m1+m2
oh so the spring balance shows 2m1m2/m1+m2 since it divides the tension produced by g to get the mass
that's why two teams pulling a rope with the same force of 1000N will produce a tension of 1000N !

because tension=2*1000*1000/(1000+1000)=1000N
wow now i am so clear about this :)
 
  • #35
Suyash Singh said:
force=2m1m2/m1+m2
You missed out g and you should use parentheses in the denominator, but other than that, yes.
 
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1. What is a spring balance and how does it work?

A spring balance is a simple measuring device used to determine the weight of an object. It works by utilizing Hooke's law, which states that the force required to extend or compress a spring is directly proportional to the distance it is stretched or compressed. The spring balance consists of a spring attached to a hook on one end and a scale on the other. When an object is suspended from the hook, the spring stretches and the scale indicates the weight of the object.

2. What is the difference between a spring balance and a regular scale?

The main difference between a spring balance and a regular scale is the mechanism used to measure weight. A regular scale typically uses a system of levers and counterweights, while a spring balance uses the principle of Hooke's law. Spring balances are also more portable and compact compared to regular scales.

3. How accurate is a spring balance?

The accuracy of a spring balance depends on the quality of the spring used and the design of the device. Generally, spring balances have a margin of error of around 1-2%. However, this can vary depending on the weight and size of the object being measured.

4. Can a spring balance be used to measure other forces besides weight?

Yes, a spring balance can be used to measure other forces such as tension and compression. This is because the force required to extend or compress a spring is directly proportional to the distance it is stretched or compressed. However, the scale on a spring balance is calibrated to measure weight, so the readings may not be accurate for other forces.

5. How do I read the measurements on a spring balance?

The scale on a spring balance is typically marked with evenly spaced increments, usually in grams or Newtons. To read the measurement, simply note the number on the scale where the spring ends. This indicates the weight of the object in grams or Newtons.

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