How Does a Spring Scale Measure the Total Force in a Balanced Meter Ruler Setup?

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The discussion centers on the calculation of the reading on a spring scale in a balanced meter ruler setup, where a 1.2N ruler is suspended with an apple and a weight. The initial answer suggests the reading is 1.75N, combining the weights of the apple and the additional weight. However, participants express confusion, arguing that the weight of the meter ruler itself, which is 1.2N, should also be considered in the total force measured by the spring scale. They conclude that the correct reading should be 2.95N, as the scale must account for the ruler's weight in a stable, balanced system. The problem's lack of clarity regarding the scale's calibration is also noted.
xiaomeizi
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Homework Statement


A uniform metre ruler of weight of 1.2N suspended at its mid-point from a spring scale which is calibrated in Newton(N).
An apple with weight of 0.75N, suspended from the metre rule at the 10cm mark, is balance by a weight of 1N, which is suspended from the 80cm mark. Assume g=10N/kg.
Q. what is the reading on the spring balance?

The reading on the spring balance= weight of both apple and the 100N. (0.75N+1N=1.75N)
I don't understand the answer. what about the weight of the metre ruler, balanced out by the upwards force ? I thought the spring scale is used to measure the tension, if that is the case, the tension should include the weight of the metre rule, isn't it ? could anyone dismiss my doubts? Thanks you. :D
 
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Homework Statement


A uniform metre ruler of weight of 1.2N suspended at its mid-point from a spring scale which is calibrated in Newton(N).
An apple with weight of 0.75N, suspended from the metre rule at the 10cm mark, is balance by a weight of 1N, which is suspended from the 80cm mark. Assume g=10N/kg.
Q. what is the reading on the spring balance?



Homework Equations





The Attempt at a Solution



The reading on the spring balance= weight downwards due to gravity
the answer suggested that (0.75N+1N=1.75N)
BUT, I don't understand the answer. what about the weight of the metre ruler, balanced out by the upwards force ? I thought the spring scale is used to measure the tension, if that is the case, the tension should include the weight of the metre rule, isn't it ? could anyone dismiss my doubts? Thanks you. :D
 
Yes, you are correct, the scale should read 2.95 N. Perhaps the scale was precalibrated to 0 with the meter stick alone on it? Like when you weigh a bunch of potatoes on a spring scale in a food store, the scale is adjusted to 0 with the pan weight alone on it, so you don't get overpriced. But the problem is not clear in that regard.
 


hi xiao

yes, the answer seems wrong. since the ruler is neither moving nor rotating, we have to use
the stability condition for the force. when we do that , the answer given seems wrong as it neglects the weight of the ruler...
 


IssacNewton said:
hi xiao

yes, the answer seems wrong. since the ruler is neither moving nor rotating, we have to use
the stability condition for the force. when we do that , the answer given seems wrong as it neglects the weight of the ruler...

Thanks for your reply.
 
(two threads merged)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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