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

[Suyash Singh]

Homework Statement

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.

 

[Chandra Prayaga]

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?

 

[haruspex]

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?

 

[Suyash Singh]

tension in both strings connected is (m1-m2)g. :(

 

[Suyash Singh]

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

 

[Chandra Prayaga]

Why are you stating that the upward force on m₁ is m₂g?

 

[Suyash Singh]

Why are you stating that the upward force on m₁ is m₂g?

because the other mass is pulling it up

 

[Chandra Prayaga]

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?

 

[Suyash Singh]

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)

 

[Chandra Prayaga]

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?

 

[jbriggs444]

You forgot to answer this question:

Is the system static?

 

[haruspex]

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.

 

[Suyash Singh]

 

[Suyash Singh]

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

 

[haruspex]

T2=m2g
T1=m1g

No. For the third time, do not assume the system is static.

 

[haruspex]

spring balance strings have no tension because they are perpendicular

Strings have this clever trick of carrying tension around corners and over pulleys.

 

[Suyash Singh]

No. For the third time, do not assume the system is static.

t1<m1g
t2<m2g

 

[Suyash Singh]

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?

 

[Suyash Singh]

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

 

[Suyash Singh]

oh and if both masses were equal

both mass=m
then reading of spring balance would be m

 

[haruspex]

is it like that the weight of only one mass is being measured and the other mass is to hold the spring balance?

No.

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.

 

[Suyash Singh]

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)

 

[jbriggs444]

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?

 

[haruspex]

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?

 

[Suyash Singh]

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.

 

[Suyash Singh]

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

 

[jbriggs444]

1000N because they won't cancel out.

Right. So why do you think that the forces add in the situation for this problem?

 

[haruspex]

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.

 

[Suyash Singh]

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.

 

[Suyash Singh]

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.