Weighing machine in an elevator

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
Initially , let the tension in spring be T

Since hanging mass is just touching the weighing scale , T = 5g

When elevator starts to accelerate upwards with acceleration 'a' , then on applying a fictitious force '5a' downwards and doing a force balance on the mass

T + N = 5g + 5a

N is the normal force from the weighing machine and T will not change

So, N = 5a

I am getting D) I.e data insufficient ,but this is incorrect .

Given answer is B) .

What is my mistake ?

 

[Chestermiller]

I don't see any mistake in what you did. Your reasoning seems flawless to me.

 

[gneill]

The question is badly posed. You are correct that the answer should be D: Data insufficient.

Without specifying the value of the spring constant and the magnitude of the acceleration a, there is no way to know what force the weighing machine will be subjected to in relation to (5 kg)(g).

 

[Chestermiller]

The question is badly posed. You are correct that the answer should be D: Data insufficient.

Without specifying the value of the spring constant and the magnitude of the acceleration a, there is no way to know what force the weighing machine will be subjected to in relation to (5 kg)(g).

Actually, the spring constant is not needed if the displacement of the mass (relative to the elevator) is assumed negligible during the upward acceleration.

 

[Jahnavi]

Actually, the spring constant is not needed if the displacement of the mass (relative to the elevator) is assumed negligible during the upward acceleration.

I agree .

 

[gneill]

Actually, the spring constant is not needed if the displacement of the mass (relative to the elevator) is assumed negligible during the upward acceleration.

Alright I can go along with that, but I don't think there's any way to relate the specific "5 kg weight" reading to an unspecified acceleration without knowing more details about the whole scenario, including the specifics of weighing machine operation and the spring constant value.

 

[brijbihari sharma]

weighing machine measures Normal force, when there is no acceleration object was just touching the weighing machine. means restoring force in spring must be equal to object's weight. so normal would be zero at that time. what I see here is that as elevator accelerate upward weight of object increases that will slightly stretch the spring so that restoring force on spring would also increase. that would decrease normal force (provided by weighing machine ) acting on the object. so It's reading must be below 5 kg. since the more it accelerate the more it would stretch the upper spring and the more restoring force would be there since restoring force is propositional to stretch.