# Why Does the Force Meter Show Different Readings in Similar Situations?

• Iamconfused123
Iamconfused123
Homework Statement
I don't understand why is the soultion on 4th and last picture the way it is
Relevant Equations
F=ma

I don't get how is the 4th case different from the 1st case? In both cases the weights are hanging and are not accelerating, but somehow in 4th case the force meter shows 0N while in 1st shows 10N.
All other meters show 10N but the last one.
Now, I don't know hot to solve last one. I tried treating it like pulley problem with two weights. Tried to calculate the acceleration, a=((M-m)/(M+m))*g, and then M(4kg)*a(6)=F=24N. But, it's not, apparently it's 16N.
Can someone please explain this to me? Thank you.

You are well on your way to solving the last problem. However, the force you have computed is the net force on the big mass. How is this related to the force on that mass due to the string tension?

Iamconfused123
Orodruin said:
You are well on your way to solving the last problem. However, the force you have computed is the net force on the big mass. How is this related to the force on that mass due to the string tension?
ohh, (a+g)×m(1kg)=16×1kg=16N. Or Tension on big mass= mg-ma=40N-24N=16N.

Thank you.

Last edited:
Orodruin said:
You are well on your way to solving the last problem. However, the force you have computed is the net force on the big mass. How is this related to the force on that mass due to the string tension?
Can you also help me with 4th example. Why in the 4th example the force meter shows 0N, If we have weight pulling on the spring on the right side, the weight on the left side is there just to keep the housing from moving, but that does not stop the weight on the right side to pull the spring, right?

Iamconfused123 said:
Why in the 4th example the force meter shows 0N
It doesn’t. It shows 10 N.

SammyS and Iamconfused123
Orodruin said:
It doesn’t. It shows 10 N.
Thanks, then solutions are wrong.

Iamconfused123 said:
Tried to calculate the acceleration, a=((M-m)/(M+m))*g, and then M(4kg)*a(6)=F=24N. But, it's not, apparently it's 16N.
You have found the correct expression for the acceleration. $$a=\frac{M-m}{M+m}g=\frac{4-1}{4+1}\times 10~(\text{m/s}^2)=6~(\text{m/s}^2).$$ The net force on ##m## is up (positive) and is the sum of the unknown tension ##T## (up) plus the weight ##mg## (down).
##F_{\text{net}}=T +(-mg)=T-mg.##
Also, the net force on ##m## is equal to the mass times its acceleration according to Newton's second law. So
##F_{\text{net}}=T-mg = ma \implies T=m(g+a).##
Put in the numbers
##T=1~(\text{kg})\times(10+6)~(\text{m/s}^2)=16~\text{N}.##

The given answer is correct. In the last figure, no reading of the force meter is shown. Instead, we see a hint to use ##10~(\text{m/s}^2)## for the acceleration of gravity.

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