Mass and apparant weight

[Uth]

Hello everybody!

 

Can we find the mass of an object from its apparent weight, i.e. what's measured by scales?

 

Thank you in advance!

[ajb]

If you know the acceleration due to the background gravitational field, then you can.

 

Use [math]F = ma[/math], which really defines the mass as the proportionality between force and acceleration.

 

Here on the Earth's surface we have [math] a=g \approx 9.81 \: ms^{-2}[/math] meters per second per second.

[Uth]

I think we can use F = m*a, or W = m*g in case of actual weight, i.e. gravitational force.

Isn't apparent weight different?

[ajb]

They are the same unless the there is some other force acting.

[D H]

Apparent weight is the net force acting on a body less the actual weight, or all of the sum of all forces acting on a body except for gravity. The only forces acting on an object on a spring scale are the normal force and gravity. In other words, the apparent weight of an object sitting on a spring scale is the normal force.

 

Can one use F=m*a to calculate the mass given the apparent weight? Of course. All you need to know is the acceleration due to this force. The acceleration due to gravity can only be computed; there is no way to measure it. The acceleration due to apparent weight, on the other hand, is exactly what accelerometers and gravimeters measure. If you want a precise estimate of the mass you need a precise measurement of apparent weight and a precise gravimeter reading. If less precision is needed, look up the apparent acceleration in your location. If even less precision is needed, use the standard acceleration due to gravity, 9.80665 m/s².

[traveler]

Can one use F=m*a to calculate the mass given the apparent weight? Of course. All you need to know is the acceleration due to this force.

 

 

F=m*a when an object is sitting on a scale?

 

F=zero because a=zero

 

or is the acceleration of the object 32.174 ft/sec^2??? Prove it!

[D H]

Prove it!

I will work in the planet-fixed (i.e., rotating) reference frame. Because this is a rotating frame, a fictitious centrifugal force is needed to make Newton's second law applicable in this frame:

 

[math]\mathbf F_{\text{eff}} = \mathbf F_{\text{inertial}} -

m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf r)[/math]

 

I have left out the Coriolis and rotational acceleration terms because (a) the object is stationary and this has no Coriolis acceleration, and (b) the Earth's rotational acceleration is very, very small and safely can be ignored.

 

The real forces (I am assuming Newtonian mechanics, where gravity is real force) acting on the object are gravity and the normal force. As the object is stationary in the planet-fixed frame, the effective force on the object is identically zero. Putting these together,

 

[math]\mathbf F_{\text{eff}} = 0 = \mathbf F_{\text{grav}} + \mathbf F_{\text{normal}} -

m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf r)[/math]

 

or

 

[math]\mathbf F_{\text{normal}} = -(\mathbf F_{\text{grav}} -

m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf r))[/math]

 

This is what a spring scale measures. (Almost. A spring scale actually measures the displacement of a spring, and the displacement of the provides the normal force in question.)

 

Define the apparent acceleration [math]\mathbf a_{\text{apparent}}[/math] as the acceleration as measured in the rotating frame that would result if the normal force were not present. The only real force besides the normal force is gravity. The apparent force (observer fixed with respect to the rotating frame) in the absence of the normal force is

 

[math]\mathbf F_{\text{apparent}} = \mathbf F_{\text{grav}} -

m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf r)[/math]

 

Dividing by the mass yields the apparent acceleration:

 

 

[math]\mathbf a_{\text{apparent}} \equiv \frac{1}{m}\,(\mathbf F_{\text{grav}} -

m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf r))[/math]

 

Applying this term to the expression for the normal force,

 

[math]\mathbf F_{\text{normal}} = -m\,\mathbf a_{\text{apparent}}[/math]

 

or

 

 

[math]

m = \frac{||\mathbf F_{\text{normal}}||}{||\mathbf a_{\text{apparent}}||}

[/math]

 

This might look like sophistry in that appears that I have rather arbitrary defined an acceleration that lets me compute the mass given the normal force. This is not sophistry. The apparent acceleration is exactly the quantity that accelerometers and gravimeters measure.

[traveler]

So how far does the object travel in 2 seconds towards the center of the Earth when on the scale?

[D H]

Are you being intentionally thick?

[traveler]

Are you being intentionally thick?

 

No, I'm just wondering how you can say an object on a scale has an acceleration towards the center of the Earth?

 

Is the object getting closer to the center of the Earth, or further away? What are you basing the acceleration on? Is the object increasing or decreasing velocity while on a scale on the Earth, compared to the center of the Earth? If not, is not the acceleration ZERO?

[insane_alien]

No, I'm just wondering how you can say an object on a scale has an acceleration towards the center of the Earth?

 

Is the object getting closer to the center of the Earth, or further away? What are you basing the acceleration on? Is the object increasing or decreasing velocity while on a scale on the Earth, compared to the center of the Earth? If not, is not the acceleration ZERO?

 

oh ffs, you measure the force required to cancel out the acceleration. this force is equal to mass time acceleration due to gravity in scalar form.

[traveler]

oh ffs, you measure the force required to cancel out the acceleration. this force is equal to mass time acceleration due to gravity in scalar form.

 

So the acceleration is actually zero, correct?

[Kyrisch]

The existence of force is not dependent on acceleration, rather one can calculate the force exerted on a body by multiplying its mass by the amount it is being accelerated. Since an object will accelerate at g in free fall, in order to keep the object at rest an opposing force equal in magnitude to act upon it. If it would otherwise accelerate at g, then the force acting upon it is mg. If it is at rest, an opposing force of mg is being applied. Therefore, knowing the weight (one definition of which is the force required to keep the object at rest in a gravitational field), one can calculate the mass.

 

So the acceleration is actually zero, correct?

 

Yes, the object is obviously undergoing an acceleration of zero magnitude, that's what being in equilibrium (no net forces acting upon it) MEANS. If it were accelerating, the scale would give a false reading.

[D H]

this force is equal to mass time acceleration due to gravity in scalar form.

Not quite. The planet is rotating. Think about it this way: If the Earth were rotating 17 times faster than it is now, objects at the equator would fly off into orbit. In other words, their apparent weight would become zero.

 

Another way to look at it: Astronauts in the Shuttle in orbit around the Earth do have near-zero apparent weight. (That's why we call it "weightlessness".) On the other hand, their actual weight, tautologically defined as mass times the acceleration due to gravity, is about 10% or so smaller than their actual weight on the surface of the Earth.

 

While the Earth is rather unlikely to rotating 17 times than it is now, the same is not true for asteroids. One side effect of solar radiation pressure is that radiation pressure can induce a torque on the asteroid. This second-order effect is called the Yarkovsky-O'Keefe-Radzievskii-Paddack effect, or YORP effect for short. Under the right conditions the Sun can slowly make an asteroid spin faster and faster. The Sun can make an asteroid throw itself apart! Two articles in Science on 1999 KW₄:

S.J. Ostro et. al., "Radar Imaging of Binary Near-Earth Asteroid (66391) 1999 KW4," Science 314(5803), 1276-1280 (24 Nov 2006) http://echo.jpl.nasa.gov/~ostro/kw4/ostro+kw4.pdf.

From the article,

Together, Alpha's size, shape, spin, density, and porosity reveal it to be an unconsolidated gravitational aggregate close to its breakup spin rate, suggesting that KW4’s origin involved spin-up and disruptive mass shedding of a loosely bound precursor object. The disruption may have been caused by tidal effects of a close encounter with a planet or by torques due to anisotropic thermal radiation of absorbed sunlight (the YORP effect).

Also see http://echo.jpl.nasa.gov/~ostro/kw4/index.html.

 

S.C. Lowry, et. al. "Direct detection of the asteroidal YORP effect," Science 316 (5822), 272-4 (13 Apr 2007)

From the article,

The Yarkovsky-O'Keefe-Radzievskii-Paddack (YORP) effect is believed to alter the spin states of small bodies in the solar system. However, evidence for the effect has so far been indirect. We report precise optical photometric observations of a small near-Earth asteroid, (54509) 2000 PH5, acquired over 4 years. We found that the asteroid has been continuously increasing its rotation rate. We simulated the asteroid's close Earth approaches from 2001 to 2005, showing that gravitational torques cannot explain the observed spin rate increase.

[Uth]

So, if the object isn't accelerating and it's perpendicular to the surface, the apparent and actual weight are equal in magnitude.

[D H]

So, if the object isn't accelerating and it's perpendicular to the surface, the apparent and actual weight are equal in magnitude.

Accelerating with respect to what? If the object isn't accelerating with respect to the surface, it depends on whether the gravitational body is rotating. For example, apparent and actual weight are not equal-but-opposite on the Earth.

 

The orientation of the object is irrelevant, so long as the object is touching the surface.

[Uth]

Maybe the difference between them (actual and apparent weight) is minor on the earth?

 

Accelerating with respect to what? If the object isn't accelerating with respect to the surface, it depends on whether the gravitational body is rotating. For example, apparent and actual weight are not equal-but-opposite on the Earth.

 

The orientation of the object is irrelevant, so long as the object is touching the surface.

 

What about the following example from Wikipedia

 

"

In a simple case such as a 40 kg object resting upon a table, the normal force on the object is equal but in opposite direction to the gravitational force applied on the object i.e. the weight of the object. In this case the normal force is given by, 40 kg · 9.81 m/s2=392.4 newtons where 9.81 m/s2 is equal to the acceleration due to gravity (near the Earth's surface).

 

In another case where the same object as mentioned above is on a 40 degree incline, we have to insert cos θ into the equation for normal force. Fnormal = mass · gravity · cos θ. So solving for the normal force, we get: FN = 40kg · 9.81m/s2 · cos 40° = 300.6 newtons

"

http://en.wikipedia.org/wiki/Normal_force