Reading on a Scale in an Elevator

e-zero

Homework Argument

A 65-kg woman descends in an elevator that briefly acclerates at 0.20g downwardly when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the calibration read?

Homework Equations

ƩF = ma
mg - Fn = 1000(0.20g)

The Attempt at a Solution

I have an reply that states that the calibration needs to exert a forcefulness of 0.80mg which will give a reading of 0.80m = 52kg, but I do not understand why this makes sense.

Mandelbroth

Homework Statement

A 65-kg woman descends in an elevator that briefly acclerates at 0.20g downwards when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read?

Homework Equations

ƩF = ma
mg - Fn = m(0.20g)

The Attempt at a Solution

I have an reply that states that the scale needs to exert a force of 0.80mg which volition give a reading of 0.80m = 52kg, but I do not understand why this makes sense.

What doesn't make sense about it?

e-zero

why do we but eliminate 'g' to find the calibration reading?

PhanthomJay

That is considering the calibration is calibrated in kilograms which is a mass unit of measurement and not a force unit. 52 kg of mass is 52g or 520 Newtons of force (weight) using 1000 = 10 thousand/sec/sec.

e-zero

Ok. Can you lot as well explain why the formula is mg - Fn = m(0.20g) and Non Fn - mg = thousand(0.20g) ??

CAF123

Ok. Can you also explain why the formula is mg - Fn = g(0.20g) and NOT Fn - mg = one thousand(0.20g) ??

The acceleration of the elevator is downward so it is convenient to ready the 1D coordinate system such that downwardly is the positive direction. This leads to the eqn mg - Fn = 0.20mg. Alternatively, if you accept downward as negative, then you have Fn - mg = -0.20mg which is equivalent.

e-zero

Ok so if we utilise mg - Fn = m(0.20g) and we solve for Fn we will get:

Fn = 0.80mg

Since nosotros took downwards to be the positive management and then Fn = 0.80mg will be downwards since it'south positive. I know I'm wrong here, but mathematically I cannot figure out why.

CAF123

Ok and so if we utilise mg - Fn = m(0.20g) and we solve for Fn nosotros will get:
Fn = 0.80mg

Yeah

Since we took downwards to exist the positive direction and then Fn = 0.80mg will be downwards since it'due south positive. I know I'm incorrect here, only mathematically I cannot effigy out why.

You expect a positive number hither. By setting up the eqn in the first identify, y'all knew that the normal forcefulness was directed in the negative direction in this case. i.e when doing a costless body diagram you identify all the forces then employ NII.

The vector equation is ##m \underline{a} = mg \chapeau{y} - F_n \chapeau{y}## from which you excerpt the scalar equation ##ma_y = mg - F_n##. ##F_n## is the magnitude of the vector ##F_n \hat{y}## and, equally such, is not negative.

e-zero

I'm just confused cause when we used the equations for mechanics (distance, speed, time) our concluding result would be either positive or negative which would indicate the direction, but now that we are using Newton'south equation it seems to non be the case.

Is this because with Newton'southward equations we are, in a style, creating the equation before we summate? as opposed to the mechanic equations which, in a fashion, were always static.

tiny-tim

hi e-zero!

(try using the X₂ button just above the Answer box )

Ok so if we utilize mg - Fn = m(0.20g) …
Since we took down to be the positive direction then Fn = 0.80mg volition be downwards since information technology's positive.

nope

when you lot wrote mg - F_n, y'all were already assuming that F_n was upwardly

if your F_n was downward, you would have written mg + F_n = k(0.20g),

giving y'all F_n = -m(0.80g),

proving that your normal forcefulness was -m(0.80g) downward, which is m(0.80g) upward!!